ML Aggarwal Class 9 Maths Solutions Chapter 15 Circle

Exercise 15.1

Question 1. Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.
Answer: Given - Base = 6 cm, Height = 4 cm. Using the formula, Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \), we get Area = \( \frac{1}{2} \times 6 \times 4 = 12 \) cm².
In simple words: To find the area, multiply the base and height, then divide by 2. The area is 12 cm².

Exam Tip: Always use the formula \( \frac{1}{2} \times \text{base} \times \text{height} \) for triangles when you know the base and height directly.

Question 2(i). Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm.
Answer: Let a = 3 cm, b = 4 cm, c = 5 cm. First, calculate the semi-perimeter: \( s = \frac{a + b + c}{2} = \frac{3 + 4 + 5}{2} = \frac{12}{2} = 6 \) cm. Using Heron's formula: Area = \( \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \) cm².
In simple words: When you know all three sides, use Heron's formula. First find half the perimeter (s), then apply the square root formula. The area is 6 cm².

Exam Tip: Heron's formula works for any triangle when all three sides are known - no need for height or angles.

Question 2(ii). Find the area of a triangle whose sides are 29 cm, 20 cm and 21 cm.
Answer: Let a = 29 cm, b = 20 cm, c = 21 cm. Calculate semi-perimeter: \( s = \frac{29 + 20 + 21}{2} = \frac{70}{2} = 35 \) cm. Applying Heron's formula: Area = \( \sqrt{35(35-29)(35-20)(35-21)} = \sqrt{35 \times 6 \times 15 \times 14} = \sqrt{44100} = 210 \) cm².
In simple words: Find half the perimeter, which is 35 cm. Then multiply the differences and take the square root to get 210 cm².

Exam Tip: Be careful with large numbers - factor them carefully before taking the square root to simplify the calculation.

Question 2(iii). Find the area of a triangle whose sides are 12 cm, 9.6 cm and 7.2 cm.
Answer: Let a = 12 cm, b = 9.6 cm, c = 7.2 cm. Find semi-perimeter: \( s = \frac{12 + 9.6 + 7.2}{2} = \frac{28.8}{2} = 14.4 \) cm. Using Heron's formula: Area = \( \sqrt{14.4(14.4-12)(14.4-9.6)(14.4-7.2)} = \sqrt{14.4 \times 2.4 \times 4.8 \times 7.2} = \sqrt{1194.39} = 34.56 \) cm².
In simple words: Calculate the semi-perimeter as 14.4 cm. Then use Heron's formula to get an area of 34.56 cm².

Exam Tip: When dealing with decimal side lengths, maintain precision throughout the calculation - rounding too early leads to incorrect final answers.

Question 3. Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.
Answer: Let a = 34 cm, b = 20 cm, c = 42 cm. Semi-perimeter: \( s = \frac{34 + 20 + 42}{2} = \frac{96}{2} = 48 \) cm. Area = \( \sqrt{48(48-34)(48-20)(48-42)} = \sqrt{48 \times 14 \times 28 \times 6} = \sqrt{112896} = 336 \) cm². The shortest side is 20 cm. Using Area = \( \frac{1}{2} \times \text{base} \times \text{height} \): \( 336 = \frac{1}{2} \times 20 \times h \), so \( h = \frac{336 \times 2}{20} = \frac{672}{20} = 33.6 \) cm.
In simple words: First find the triangle's area using Heron's formula, which is 336 cm². Then use the base-height formula with the shortest side to find the altitude is 33.6 cm.

Exam Tip: Remember that the same area can be expressed using any side as the base - rearrange the formula to find the corresponding height.

Question 4. The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate of Rs 10 lakh per hectare, find its selling price. [1 hectare = 10000 m²]
Answer: Let a = 975 m, b = 1050 m, c = 1125 m. Semi-perimeter: \( s = \frac{975 + 1050 + 1125}{2} = \frac{3150}{2} = 1575 \) m. Area = \( \sqrt{1575(1575-975)(1575-1050)(1575-1125)} = \sqrt{1575 \times 600 \times 525 \times 450} \). Factoring: \( = \sqrt{525^2 \times 150^2 \times 2^2 \times 3^2} = 525 \times 150 \times 2 \times 3 = 472500 \) m². Converting to hectares: \( \frac{472500}{10000} = 47.25 \) hectares. Selling price = \( 47.25 \times \text{Rs } 10,00,000 = \text{Rs } 4,72,50,000 \).
In simple words: Find the field's area using Heron's formula to get 472500 m². Convert this to 47.25 hectares. Multiply by the price per hectare to get the total selling price of Rs 4,72,50,000.

Exam Tip: Watch unit conversions carefully - convert m² to hectares before calculating the final price. Working with the correct units throughout prevents calculation errors.

Question 5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.
Answer: Given - triangle ABC with base BC = 12 cm and hypotenuse AC = 13 cm. Using Pythagoras' theorem: \( AC^2 = AB^2 + BC^2 \), so \( 13^2 = AB^2 + 12^2 \), which gives \( 169 = AB^2 + 144 \), therefore \( AB^2 = 25 \) and \( AB = 5 \) cm. Area = \( \frac{1}{2} \times 12 \times 5 = 30 \) cm². Perimeter = \( 5 + 12 + 13 = 30 \) cm.
In simple words: Find the third side using Pythagoras' theorem - it comes out to 5 cm. Then calculate the area as 30 cm² and the perimeter as 30 cm.

Exam Tip: In a right-angled triangle, the two sides forming the right angle are the base and height - use them directly in the area formula without needing Heron's formula.

Question 6. Find the area of an equilateral triangle whose side is 8 m. Give your answer correct to two decimal places.
Answer: Given - side = 8 m. For an equilateral triangle, Area = \( \frac{\sqrt{3}}{4} \times (\text{side})^2 = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} = 16 \times 1.732 = 27.71 \) m².
In simple words: For an equilateral triangle, use the special formula with \( \sqrt{3} \). Substituting 8 gives approximately 27.71 m².

Exam Tip: Remember the equilateral triangle formula \( \frac{\sqrt{3}}{4} \times (\text{side})^2 \) - it's faster than Heron's formula for equilateral triangles. Always use \( \sqrt{3} \approx 1.732 \) for decimal approximations.

Question 7. If the area of an equilateral triangle is 81\(\sqrt{3}\) cm², find its perimeter.
Answer: Given - Area = \( 81\sqrt{3} \) cm². Using the equilateral triangle area formula: \( 81\sqrt{3} = \frac{\sqrt{3}}{4} \times (\text{side})^2 \). Multiplying both sides by 4 and dividing by \( \sqrt{3} \): \( (\text{side})^2 = \frac{81\sqrt{3} \times 4}{\sqrt{3}} = 81 \times 4 = 324 \), so side = \( \sqrt{324} = 18 \) cm. Perimeter = \( 3 \times 18 = 54 \) cm.
In simple words: Work backwards from the area formula to find the side length is 18 cm. Then multiply by 3 to get the perimeter of 54 cm.

Exam Tip: When given area and asked to find the side, reverse the formula - rearrange algebraically to isolate the side term before taking the square root.

Question 8. If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.
Answer: Given - Perimeter = 36 cm. Since Perimeter = \( 3 \times \text{side} \), we have side = \( \frac{36}{3} = 12 \) cm. Area = \( \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} = 36 \times 1.732 = 62.4 \) cm². For height, in an equilateral triangle the altitude bisects the base, creating a right triangle. Using Pythagoras' theorem: \( 12^2 = h^2 + 6^2 \), so \( 144 = h^2 + 36 \), giving \( h^2 = 108 \) and \( h = \sqrt{108} = 10.4 \) cm.
In simple words: The side is 12 cm. The area comes to 62.4 cm² using the equilateral formula. The height is found using Pythagoras' theorem as 10.4 cm.

Exam Tip: In an equilateral triangle, the altitude from any vertex to the opposite side creates two congruent right triangles - use this property with Pythagoras' theorem to find height.

Question 9(i). If the lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 48 cm, find its area.
Answer: Let the sides be a = 3x, b = 4x, c = 5x. Since perimeter = 48 cm: \( 3x + 4x + 5x = 48 \), so \( 12x = 48 \) and \( x = 4 \). Therefore, a = 12 cm, b = 16 cm, c = 20 cm. Semi-perimeter: \( s = \frac{12 + 16 + 20}{2} = 24 \) cm. Area = \( \sqrt{24(24-12)(24-16)(24-20)} = \sqrt{24 \times 12 \times 8 \times 4} = \sqrt{9216} = 96 \) cm².
In simple words: Use the ratio to express sides in terms of x. Solve for x using the perimeter equation to get x = 4. Then apply Heron's formula with the actual side lengths to find the area is 96 cm².

Exam Tip: When sides are in a given ratio, always express them as multiples of a common variable - this simplifies the perimeter equation and makes finding actual lengths straightforward.

Question 9(ii). The sides of a triangular plot are in the ratio 3 - 5 - 7 and its perimeter is 300 m. Find its area. Take \( \sqrt{3} = 1.732 \).
Answer: Let the sides be represented as 3x, 5x, and 7x. Since the perimeter is 300 m, we have 3x + 5x + 7x = 300, which gives 15x = 300, so x = 20. Therefore, the three sides measure 60 m, 100 m, and 140 m. The semi-perimeter is 150 m. Using Heron's formula:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6750000} = 2598.07 \approx 2598 \text{ m}^2 \]
In simple words: Find the length of each side using the given ratio and perimeter. Then use Heron's formula with the semi-perimeter to get the area.

Exam Tip: Always compute the semi-perimeter first, then apply Heron's formula step-by-step with careful arithmetic to avoid calculation errors.

Question 10. ABC is a triangle in which AB = AC = 4 cm and \( \angle A = 90° \). Calculate the area of \( \triangle ABC \). Also find the length of perpendicular from A to BC.
Answer: Given AB = AC = 4 cm and angle A is 90 degrees. Since two sides meeting at the right angle are equal, this is a right isosceles triangle. The area can be found using the formula \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \text{ cm}^2 \). To find BC, apply the Pythagoras theorem: \( BC^2 = 4^2 + 4^2 = 32 \), so \( BC = 4\sqrt{2} \) cm. Now, if h is the perpendicular from A to BC, then \( 8 = \frac{1}{2} \times 4\sqrt{2} \times h \), which gives \( h = \frac{8}{2\sqrt{2}} = 2\sqrt{2} = 2.83 \) cm.
In simple words: For a right-angled triangle, multiply the two sides that touch the right angle and divide by 2 to get the area. To find the perpendicular height to the longest side, use the area formula again, but this time solve for the height.

Exam Tip: When finding the perpendicular to the hypotenuse, set up the equation Area = \( \frac{1}{2} \times \text{hypotenuse} \times h \) and solve for h; this is a common exam pattern.

Question 11. Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.
Answer: In the isosceles triangle, let AB = AC = 12 cm be the equal sides. Since the perimeter is 30 cm, the base BC equals 30 - 12 - 12 = 6 cm. The semi-perimeter is s = 15 cm. For an isosceles triangle with equal sides a and base b, the area formula is \( \frac{1}{4}b\sqrt{4a^2 - b^2} \). Substituting the values:
\[ A = \frac{1}{4} \times 6 \times \sqrt{4(12)^2 - 6^2} = \frac{1}{4} \times 6 \times \sqrt{576 - 36} = \frac{1}{4} \times 6 \times \sqrt{540} = \frac{1}{4} \times 6 \times 23.24 = 34.86 \text{ cm}^2 \]
In simple words: In an isosceles triangle, once you know the two equal sides and perimeter, find the third side. Then plug these measurements into the special isosceles triangle area formula.

Exam Tip: Remember the compact isosceles formula \( \frac{1}{4}b\sqrt{4a^2 - b^2} \) - it is faster than Heron's formula for these problems and reduces arithmetic errors.

Question 12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.
Answer: Let AB = AC = x cm represent the two equal sides, and BC = 6 cm is the base. From the perimeter condition, x + x + 6 = 16, so 2x = 10 and x = 5 cm. Now apply the isosceles triangle area formula with a = 5 and b = 6:
\[ A = \frac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2} = \frac{1}{4} \times 6 \times \sqrt{100 - 36} = \frac{1}{4} \times 6 \times \sqrt{64} = \frac{1}{4} \times 6 \times 8 = 12 \text{ cm}^2 \]
In simple words: Use the perimeter to find the length of each equal side. Then use those lengths in the isosceles area formula.

Exam Tip: Always set up a variable equation from the perimeter first to determine all three side lengths before applying any area formula.

Question 13. The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm\( ^2 \).
Answer: The two sides that form the right angle are 5x cm and (3x - 1) cm. The area of a right triangle is \( \frac{1}{2} \times 5x \times (3x - 1) = 60 \). Expanding: 120 = 5x(3x - 1) = 15x\( ^2 \) - 5x, which gives 15x\( ^2 \) - 5x - 120 = 0. Dividing by 5: 3x\( ^2 \) - x - 24 = 0. Factoring: (3x + 8)(x - 3) = 0, so x = 3 (since x cannot be negative). Thus AB = 5(3) = 15 cm and BC = 3(3) - 1 = 8 cm. By Pythagoras' theorem, AC\( ^2 \) = 15\( ^2 \) + 8\( ^2 \) = 225 + 64 = 289 = 17\( ^2 \), so AC = 17 cm.
In simple words: Set up an area equation with the given expressions, solve for x, then use Pythagoras' theorem to find the hypotenuse.

Exam Tip: After solving the quadratic, always check that your x value produces positive side lengths and verify using the area formula before finding the hypotenuse.

Question 14. In \( \triangle ABC \), \( \angle B = 90° \), AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the \( \triangle ABC \) is 60 cm\( ^2 \), find its perimeter.
Answer: The area of the right triangle is \( \frac{1}{2} \times BC \times AB = \frac{1}{2}(x + 1)(2x + 1) = 60 \). Simplifying: 120 = (2x + 1)(x + 1) = 2x\( ^2 \) + 3x + 1. This gives 2x\( ^2 \) + 3x - 119 = 0. Factoring: (x - 7)(2x + 17) = 0, so x = 7 (since x must be positive). Therefore AB = 2(7) + 1 = 15 cm and BC = 7 + 1 = 8 cm. Using Pythagoras' theorem: AC\( ^2 \) = 15\( ^2 \) + 8\( ^2 \) = 225 + 64 = 289, so AC = 17 cm. The perimeter is 15 + 8 + 17 = 40 cm.
In simple words: Write the area formula with the algebraic expressions given, solve for the unknown, then calculate all side lengths and add them to get the perimeter.

Exam Tip: After finding x, always substitute back into the original expressions and verify the area before computing the perimeter to confirm your solution.

Question 15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.
Answer: Let the two sides forming the right angle be x cm and (35 - x) cm, where 35 comes from 60 - 25. By Pythagoras' theorem, 25\( ^2 \) = x\( ^2 \) + (35 - x)\( ^2 \). Expanding: 625 = x\( ^2 \) + 1225 - 70x + x\( ^2 \) = 2x\( ^2 \) - 70x + 1225. Rearranging: 2x\( ^2 \) - 70x + 600 = 0, or x\( ^2 \) - 35x + 300 = 0. Solving: \( x = \frac{35 \pm \sqrt{1225 - 1200}}{2} = \frac{35 \pm \sqrt{25}}{2} = \frac{35 \pm 5}{2} \). Thus x = 20 or x = 15. If x = 20, then the other side is 15 cm. The area is \( \frac{1}{2} \times 20 \times 15 = 150 \text{ cm}^2 \).
In simple words: Use the perimeter to express one side in terms of the other, apply Pythagoras' theorem to set up an equation, solve it, then compute the area.

Exam Tip: When given perimeter and hypotenuse, remember that the sum of the two legs equals (perimeter - hypotenuse). Use this directly in Pythagoras' theorem to set up your equation efficiently.

Question 16. The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.
Answer: Let the length of the two equal sides be x cm. The base measures \( \frac{2}{3}(x + x) = \frac{4x}{3} \) cm. Since the perimeter is 40 cm:

\( x + x + \frac{4x}{3} = 40 \)

\( 2x + \frac{4x}{3} = 40 \)

\( \frac{6x + 4x}{3} = 40 \)

\( 10x = 120 \)

\( x = 12 \) cm

Therefore, the base = \( \frac{4 \times 12}{3} = 16 \) cm. The length of the two equal sides is 12 cm each, and the base is 16 cm.
In simple words: Add the three sides (two equal sides plus the base) to get 40 cm. This helps you find that each equal side is 12 cm and the base is 16 cm.

Exam Tip: Always express the base in terms of the equal sides first using the given relationship, then set up the perimeter equation to solve for the unknown side length.

Question 17. If the area of an isosceles triangle is 60 cm² and the length of each of its equal sides is 13 cm, find its base.
Answer: Let a = 13 cm (equal sides) and b = base (unknown). The area formula for an isosceles triangle is \( A = \frac{1}{4}b\sqrt{4a^2 - b^2} \).

Substituting the values:

\( 60 = \frac{1}{4}b\sqrt{4(13)^2 - b^2} \)

\( 240 = b\sqrt{676 - b^2} \)

Squaring both sides:

\( 57600 = b^2(676 - b^2) \)

\( 676b^2 - b^4 = 57600 \)

\( b^4 - 676b^2 + 57600 = 0 \)

\( b^4 - 576b^2 - 100b^2 + 57600 = 0 \)

\( b^2(b^2 - 576) - 100(b^2 - 576) = 0 \)

\( (b^2 - 100)(b^2 - 576) = 0 \)

\( b^2 = 100 \) or \( b^2 = 576 \)

\( b = 10 \) cm or \( b = 24 \) cm

Therefore, the base measures either 10 cm or 24 cm.
In simple words: Use the area formula backwards to find the base. You get two possible answers because the quadratic equation has two solutions.

Exam Tip: When solving quadratic equations from the area formula, both mathematical solutions are typically valid unless the triangle becomes impossible (check the triangle inequality theorem if needed).

Question 18. The base of a triangular field is 3 times its height. If the cost of cultivating the field at the rate of Rs. 25 per 100 m² is Rs. 60000, find its base and height.
Answer: First, find the total area. If Rs. 25 cultivates 100 m², then Rs. 60000 cultivates:

\( \text{Area} = \frac{100 \times 60000}{25} = 240000 \text{ m}^2 \)

Let the height be h meters. Then the base is 3h meters. Using the triangle area formula:

\( 240000 = \frac{1}{2} \times 3h \times h \)

\( 240000 = \frac{3h^2}{2} \)

\( h^2 = \frac{240000 \times 2}{3} = 160000 \)

\( h = \sqrt{160000} = 400 \text{ m} \)

Therefore, height = 400 m and base = 3 × 400 = 1200 m.
In simple words: Work backwards from the cost to find the area. Then use the area formula to find the height, and multiply by 3 to get the base.

Exam Tip: Always start by converting the cost information into total area, then apply the triangle area formula with the given relationship between base and height.

Question 19. A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the adjoining figure). A gardener Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs. 200 per metre leaving a space 3 m wide for a gate on one side.
Answer: The perimeter of triangle ABC is 120 + 80 + 50 = 250 m. Since a 3 m wide gate opening is left, the fencing length needed is 250 - 3 = 247 m. Total fencing cost = 247 × 200 = Rs. 49,400.

To find the area for grass plantation, use Heron's formula. Semi-perimeter s = 250/2 = 125 m.

\( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{125(125-120)(125-80)(125-50)} \)

\( = \sqrt{125 \times 5 \times 45 \times 75} = \sqrt{(25 \times 5) \times 5 \times (5 \times 3 \times 3) \times (25 \times 3)} \)

\( = \sqrt{25^2 \times 5^2 \times 3^2 \times 5 \times 3} = 25 \times 5 \times 3 \times \sqrt{15} = 375\sqrt{15} \text{ m}^2 \)

Therefore, the area needed for plantation is \( 375\sqrt{15} \) m², and the fencing cost is Rs. 49,400.
In simple words: Find the perimeter and subtract the gate width to get the fencing length, then multiply by the rate. Use Heron's formula to find the area inside the triangle.

Exam Tip: Always remember to reduce the perimeter by the gate width before calculating fencing cost. Heron's formula is essential for finding the area when all three sides are known.

Question 20. An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (shown in the adjoining figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer: Each triangular piece has sides a = 20 cm, b = 50 cm, c = 50 cm. The semi-perimeter is:

\( s = \frac{20 + 50 + 50}{2} = 60 \text{ cm} \)

Using Heron's formula:

\( \text{Area} = \sqrt{60(60-20)(60-50)(60-50)} = \sqrt{60 \times 40 \times 10 \times 10} = \sqrt{240000} = 200\sqrt{6} \text{ cm}^2 \)

Since 10 pieces are needed with 5 pieces of each color:

\( \text{Total cloth of each color} = 5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2 \)

Therefore, \( 1000\sqrt{6} \) cm² of cloth of each color is required.
In simple words: Calculate the area of one triangle piece using Heron's formula. Since there are 5 pieces of each color, multiply the single piece area by 5 to get the total cloth needed for each color.

Exam Tip: Pay attention to how many pieces of each color are needed - divide the total pieces by 2 and multiply the single piece area accordingly.

Question 21(a). In the figure (i) given below, ABC is an equilateral triangle with each side of length 10 cm. In △BCD, ∠D = 90° and CD = 6 cm. Find the area of the shaded region. Give your answer correct to one decimal place.
Answer: For equilateral triangle ABC with side 10 cm:

\( \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} \times (10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} = 43.3 \text{ cm}^2 \)

For right-angled triangle BCD where BC = 10 cm, CD = 6 cm, and ∠D = 90°:

Using the Pythagorean theorem: \( BC^2 = BD^2 + CD^2 \)

\( 10^2 = BD^2 + 6^2 \)

\( BD^2 = 100 - 36 = 64 \)

\( BD = 8 \text{ cm} \)

Area of △BCD = \( \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2 \)

Area of shaded region = Area of △ABC - Area of △BCD = 43.3 - 24 = 19.3 cm²
In simple words: Find the area of the large equilateral triangle, then subtract the area of the smaller right triangle that sits inside it. The result is the shaded part.

Exam Tip: For equilateral triangles, always use the formula \( \frac{\sqrt{3}}{4}(\text{side})^2 \). For right triangles inside, use the Pythagorean theorem to find missing sides before calculating area.

Question 21(b). In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.
Answer: From the given information: AB = AD + DB = 3 + 4 = 7 cm, and similarly AC = 7 cm (isosceles right triangle).

In right-angled triangle ADE where AD = AE = 3 cm:

\( DE^2 = AD^2 + AE^2 = 3^2 + 3^2 = 18 \)

\( DE = 3\sqrt{2} \text{ cm} \)

Since DEFG is a rectangle, GF = DE = \( 3\sqrt{2} \) cm.

In right-angled triangle ABC:

\( BC^2 = AB^2 + AC^2 = 7^2 + 7^2 = 98 \)

\( BC = 7\sqrt{2} \text{ cm} \)

From the figure, BG + GF + FC = BC. By symmetry (since △DBG ≅ △ECF by RHS axiom), BG = FC. Let BG = FC = x.

\( x + 3\sqrt{2} + x = 7\sqrt{2} \)

\( 2x = 4\sqrt{2} \)

\( x = 2\sqrt{2} \text{ cm} \)

In right triangle DBG: \( DB^2 = BG^2 + DG^2 \)

\( 4^2 = (2\sqrt{2})^2 + DG^2 \)

\( 16 = 8 + DG^2 \)

\( DG = 2\sqrt{2} \text{ cm} \)

Area of △DBG = \( \frac{1}{2} \times 2\sqrt{2} \times 2\sqrt{2} = \frac{1}{2} \times 8 = 4 \text{ cm}^2 \)

Area of △ADE = \( \frac{1}{2} \times 3 \times 3 = 4.5 \text{ cm}^2 \)

Since △DBG ≅ △ECF, area of △ECF = 4 cm²

Total shaded area = Area of (△ADE + △DBG + △ECF) = 4.5 + 4 + 4 = 12.5 cm²
In simple words: Break the shaded region into three triangles: the small one at the top and two corner triangles at the bottom. Find each area separately using the Pythagorean theorem where needed, then add them together.

Exam Tip: Recognize congruent triangles using RHS axiom to avoid calculating the same area twice. Use the fact that isosceles right triangles have equal perpendicular sides to set up equations for unknown lengths.

Exercise 15.2

Question 1(i). Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.
Answer: Consider quadrilateral ABCD with diagonal BD = 20 cm, and perpendicular distances from A and C to this diagonal being 15 cm and 9 cm respectively.

The quadrilateral can be divided into two triangles: △ABD and △BCD, both sharing the base BD.

Area of △ABD = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 15 = 150 \text{ cm}^2 \)

Area of △BCD = \( \frac{1}{2} \times 20 \times 9 = 90 \text{ cm}^2 \)

Total area of quadrilateral ABCD = 150 + 90 = 240 cm²
In simple words: A diagonal splits a quadrilateral into two triangles. If you know the diagonal length and the heights from the other two vertices to this diagonal, you can find each triangle's area separately, then add them.

Exam Tip: When a diagonal divides a quadrilateral, always use it as the common base for both triangles. The perpendicular distances from the other vertices become the respective heights.

Question 1(ii). Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.
Answer: Take quadrilateral ABCD where diagonals AC and BD meet at point M and form right angles. From the figure, AC = 18 cm and BD = 12 cm. When a quadrilateral's diagonals cross at right angles, you can calculate its area using the formula: Area = (1/2) × d1 × d2, where d1 and d2 represent the diagonal lengths.

Plugging in the numbers:

Area of quadrilateral ABCD = (1/2) × 12 × 18

\( = 6 \times 18 \)

\( = 108 \text{ cm}^2 \)

So the area of the quadrilateral is 108 cm².
In simple words: When two diagonals cross at right angles, multiply them together and divide by 2 to find the area.

Exam Tip: Always check if the diagonals meet at right angles - this special condition allows you to use the (1/2) × d1 × d2 formula instead of more complex methods.

Question 2. Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°.
Answer: Looking at the quadrilateral field ABCD, we need to split it into two triangles using the diagonal BD.

For triangle BAD:
Since ∠A = 90°, apply the Pythagorean theorem:

\( BD^2 = AB^2 + AD^2 \)

\( BD^2 = 40^2 + 9^2 \)

\( BD^2 = 1600 + 81 = 1681 \)

\( BD = \sqrt{1681} = 41 \text{ m} \)

Area of triangle BAD = (1/2) × base × height = (1/2) × 40 × 9 = 180 m²

For triangle BDC, use Heron's formula with sides a = BD = 41 m, b = BC = 28 m, and c = CD = 15 m:

Semi-perimeter (s) = (41 + 28 + 15) / 2 = 84 / 2 = 42 m

Area of triangle BDC = \( \sqrt{42(42-41)(42-28)(42-15)} \)

\( = \sqrt{42 \times 1 \times 14 \times 27} \)

\( = \sqrt{15876} \)

\( = 126 \text{ m}^2 \)

Total area of quadrilateral ABCD = 180 + 126 = 306 m²
In simple words: Split the quadrilateral into two triangles, find the length of the diagonal using Pythagoras, then calculate each triangle's area separately using the appropriate formula.

Exam Tip: When one angle is 90°, immediately use Pythagoras to find the diagonal, then split the quadrilateral to find individual triangle areas using both right-angle and Heron's formula methods.

Question 3. Find the area of the quadrilateral ABCD in which ∠BCA = 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.
Answer: The quadrilateral ABCD consists of right-angled triangle ABC and equilateral triangle ACD.

For right-angled triangle ABC (with right angle at C):
Using the Pythagorean theorem:

\( AB^2 = AC^2 + BC^2 \)

\( 13^2 = 12^2 + BC^2 \)

\( 169 = 144 + BC^2 \)

\( BC^2 = 25 \)

\( BC = 5 \text{ cm} \)

Area of triangle BCA = (1/2) × AC × BC = (1/2) × 12 × 5 = 30 cm²

For equilateral triangle ACD with side 12 cm:
Area of an equilateral triangle = \( \frac{\sqrt{3}}{4} \times (\text{side})^2 \)

\( = \frac{\sqrt{3}}{4} \times 12^2 \)

\( = \frac{\sqrt{3}}{4} \times 144 \)

\( = 36\sqrt{3} = 62.35 \text{ cm}^2 \)

Total area of quadrilateral ABCD = 30 + 62.35 = 92.35 cm²
In simple words: Find the missing side of the right triangle using Pythagoras, calculate its area. Then use the equilateral triangle formula for the second part and add both areas together.

Exam Tip: Remember the equilateral triangle area formula: (√3/4) × side² - this saves time and avoids needing to calculate the height separately.

Question 4. Find the area of quadrilateral ABCD in which ∠B = 90°, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.
Answer: Split the quadrilateral ABCD into triangles ABC and ADC using diagonal AC.

For triangle ABC (with right angle at B):
Use the Pythagorean theorem:

\( AC^2 = AB^2 + BC^2 \)

\( AC^2 = 6^2 + 8^2 \)

\( AC^2 = 36 + 64 = 100 \)

\( AC = 10 \text{ cm} \)

Area of triangle ABC = (1/2) × AB × BC = (1/2) × 6 × 8 = 24 cm²

For triangle ADC, use Heron's formula with sides a = AD = 13 cm, b = DC = 13 cm, and c = AC = 10 cm:

Semi-perimeter (s) = (13 + 13 + 10) / 2 = 36 / 2 = 18 cm

Area of triangle ADC = \( \sqrt{18(18-13)(18-13)(18-10)} \)

\( = \sqrt{18 \times 5 \times 5 \times 8} \)

\( = \sqrt{3600} \)

\( = 60 \text{ cm}^2 \)

Total area of quadrilateral ABCD = 24 + 60 = 84 cm²
In simple words: Find the diagonal with Pythagoras from the right-angled triangle, then use that diagonal to split the quadrilateral and calculate both areas separately.

Exam Tip: When given equal sides (CD = AD), the triangle is isosceles - Heron's formula becomes useful. Always find the diagonal first when there's a right angle given.

Question 5. The perimeter of a rectangular cardboard is 96 cm; if its breadth is 18 cm, find the length and the area of the cardboard.
Answer: We know that the perimeter formula for a rectangle is: Perimeter = 2 × (length + breadth) = 96 cm

Substituting the breadth value:

\( 2(l + 18) = 96 \)

\( l + 18 = 48 \)

\( l = 30 \text{ cm} \)

Now calculate the area using: Area = length × breadth

Area = 30 × 18 = 540 cm²

Therefore, the length of the cardboard is 30 cm and its area is 540 cm².
In simple words: Use the perimeter to find the missing length, then multiply length and breadth to get the area.

Exam Tip: Always rearrange the perimeter formula carefully before substituting values - a small algebra error here leads to the wrong length and area.

Question 6. The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 594 m², find its perimeter.
Answer: Let the breadth = x metres. Then length = (x + 5) metres.

Using the area formula: Area = length × breadth

\( 594 = x(x + 5) \)

\( 594 = x^2 + 5x \)

\( x^2 + 5x - 594 = 0 \)

Factor by splitting the middle term:

\( x^2 + 27x - 22x - 594 = 0 \)

\( x(x + 27) - 22(x + 27) = 0 \)

\( (x - 22)(x + 27) = 0 \)

This gives x = 22 or x = -27.

Since length cannot be negative, x = 22 m (breadth).

Length = x + 5 = 22 + 5 = 27 m

Perimeter = 2(length + breadth) = 2(27 + 22) = 2 × 49 = 98 m
In simple words: Write the length in terms of breadth, use the area to set up a quadratic equation, solve it, and reject the negative answer.

Exam Tip: When you get two solutions from a quadratic, always reject negative values for physical dimensions like length and width - only the positive answer makes sense.

Question 7(a). The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.
Answer: The shaded region consists of a horizontal path and a vertical path that cross the rectangular field. These two paths overlap at a square in the middle.

To find the shaded area, add the two rectangular paths and subtract the overlapping square (which was counted twice):

Area of shaded portion = Area of horizontal rectangle + Area of vertical rectangle - Area of overlapping square

Horizontal rectangle = 50 × 5 = 250 m²

Vertical rectangle = 35 × 5 = 175 m²

Overlapping square = 5 × 5 = 25 m²

Total shaded area = 250 + 175 - 25 = 400 m²
In simple words: The two crossing paths create two rectangles that overlap in a square. Add both rectangles and subtract the square to avoid counting it twice.

Exam Tip: When two rectangles overlap, always subtract the overlap region - forgetting this step is a common mistake that inflates your final answer.

Question 7(b). In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.
Answer: The shaded portion is found by calculating the area of the large rectangle and subtracting the total area of the five small white squares cut out from it.

Area of large rectangle = 8 × 6 = 48 cm²

Area of five small squares (each with side 2 cm) = 5 × (2 × 2) = 5 × 4 = 20 cm²

Area of shaded portion = 48 - 20 = 28 cm²
In simple words: Find the total area of the outer rectangle, then subtract all the white squares inside it.

Exam Tip: Count the number of white squares carefully - missing even one square significantly changes your answer. Mark them off as you count.

Question 8. A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all

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Question 8. A rectangular plot is 20 m long and 14 m wide. A concrete path 2 m wide is laid all around. Find the area to be laid with grass.
Answer: Let ABCD represent the rectangular plot with length 20 m and breadth 14 m. The inner rectangular region PQRS will be covered with grass. From the diagram, the width of the path on each side is 2 m, so:

PQ = 20 - (2 × 2) = 20 - 4 = 16 m

QR = 14 - (2 × 2) = 14 - 4 = 10 m

Now, the area of the grass region is calculated using the formula for a rectangle:

Area of rectangle PQRS = length × breadth = 16 × 10 = 160 m²

Therefore, the area to be laid with grass is 160 m².
In simple words: The grass area is the inner rectangle after subtracting the path width from all sides. We subtract 2 m twice (once on each side) from the length and width, then multiply to find the area.

Exam Tip: Always subtract the path width twice (from both opposite sides) when finding inner dimensions. Verify your final area makes sense by checking that it is smaller than the outer plot area.

Question 9. The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.
(i) Find the length and the breadth of the lawn.
(ii) Hence, or otherwise, find the area of the flower-beds.
Answer:
(i) Let PQRS represent the lawn. From the diagram, the outer rectangle ABCD has dimensions 30 m (length) by 12 m (breadth). The flower beds border three sides with a width of 2 m each. Working from the figure:

QR = BC - BQ - RC = 30 - 2 - 2 = 26 m

SR = CD - DS = 12 - 2 = 10 m

Therefore, the length of the lawn PQRS is 26 m and its breadth is 10 m.

(ii) The flower bed area is found by subtracting the lawn area from the total outer area:

Area of flower beds = Area of rectangle ABCD - Area of rectangle PQRS

= (AD × DC) - (QR × SR) = (30 × 12) - (26 × 10) = 360 - 260 = 100 m²

Thus, the area of the flower beds is 100 m².
In simple words: The lawn sits inside the outer rectangle. To find the flower bed area, we take the total outer area and remove the lawn area from it.

Exam Tip: When a region borders only some sides (not all), check the diagram carefully to see which dimensions remain unchanged. Always verify that flower bed dimensions match the given border widths.

Question 10. A footpath of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m², find its width.
Answer: Let ABCD be the rectangular field with length 50 m and breadth 38 m. Let the width of the footpath be x metres. The inner rectangular region PQRS (the area not covered by the path) has dimensions:

PQ = 50 - 2x metres

QR = 38 - 2x metres

Using the relationship between areas:

Area of path = Area of rectangle ABCD - Area of rectangle PQRS

492 = (50 × 38) - [(50 - 2x)(38 - 2x)]

492 = 1900 - [1900 - 100x - 76x + 4x²]

492 = 176x - 4x²

4x² - 176x + 492 = 0

x² - 44x + 123 = 0

Factoring: x² - 41x - 3x + 123 = 0

x(x - 41) - 3(x - 41) = 0

(x - 3)(x - 41) = 0

This gives x = 3 or x = 41. Since the path width cannot exceed the field's breadth, x ≠ 41. Therefore, the width of the footpath is 3 m.
In simple words: The path area equals the outer rectangle minus the inner uncovered rectangle. We set up an equation and solve for the width using factoring.

Exam Tip: When two solutions appear, check which one is physically reasonable by comparing it to the field dimensions. The smaller, realistic value is typically the answer.

Question 11. The cost of enclosing a rectangular garden with a fence all around at the rate of Rs 150 per metre is Rs 54,000. If the length of the garden is 100 m, find the area of the garden.
Answer: Given: Length of the garden = 100 m, fence cost = Rs 150 per metre, total cost = Rs 54,000. Let breadth = x metres. The perimeter of a rectangle is 2(l + b), so:

Perimeter = 2(100 + x) = (200 + 2x) metres

Total fencing cost = Rate × Perimeter = 150(200 + 2x) = 30,000 + 300x rupees

Since the total cost is Rs 54,000:

30,000 + 300x = 54,000

300x = 24,000

x = 80 metres

Therefore, breadth of the garden = 80 m. The area of the rectangular garden = length × breadth = 100 × 80 = 8,000 m².
In simple words: We use the perimeter formula and the cost information to find the missing breadth, then calculate the area by multiplying length and breadth.

Exam Tip: Always separate the perimeter calculation from the cost calculation—perimeter tells us the total length of fencing, and cost per unit helps us find the actual perimeter value.

Question 12. A rectangular floor which measures 15 m × 8 m is to be laid with tiles measuring 50 cm × 25 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?
Answer: Let ABCD represent the rectangular floor and PQRS represent the carpet. First, convert all measurements to the same unit:

Area of floor = 15 × 8 = 120 m² = 1,200,000 cm²

Area of one tile = 50 × 25 = 1,250 cm²

Number of tiles required = 1,200,000 ÷ 1,250 = 960 tiles

For the carpet portion, a 1 m border exists on all sides, so:

Length of carpet = 15 - 1 - 1 = 13 m

Breadth of carpet = 8 - 1 - 1 = 6 m

Area of carpet = 13 × 6 = 78 m²

Uncovered floor area = 120 - 78 = 42 m²

Fraction uncovered = 42/120 = 7/20

Thus, 960 tiles are needed and 7/20 of the floor remains uncovered.
In simple words: Divide the total floor area by one tile's area to get the number of tiles. To find the uncovered fraction, subtract the carpet area from the total floor area, then divide by the total floor area.

Exam Tip: Always match units before dividing areas. Simplify final fractions to their lowest terms to earn full marks.

Question 13. The width of a rectangular room is 3/5 of its length x metres. If its perimeter is y metres, write an equation connecting x and y. Find the floor area of the room if its perimeter is 32 m.
Answer: Given: Length = x metres, width = (3/5)x metres, perimeter = y metres. Using the perimeter formula for a rectangle:

y = 2(length + width) = 2[x + (3/5)x] = 2[(5x + 3x)/5] = 2(8x/5) = 16x/5

Multiplying both sides by 5: 5y = 16x, which is the required equation connecting x and y.

When y = 32 m:

5(32) = 16x

160 = 16x

x = 10 metres (length)

Width = (3/5) × 10 = 6 metres

Floor area = length × width = 10 × 6 = 60 m²

Therefore, the equation is 5y = 16x and the floor area is 60 m².
In simple words: Express the perimeter using the given relationship between width and length. Simplify to get the equation. Then substitute the given perimeter value to find the dimensions and area.

Exam Tip: Keep the perimeter equation in its simplest form. When solving for one variable, substitute carefully and double-check your arithmetic.

Question 14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.
Answer: Let ABCD be the rectangular garden with length 10 m and breadth 16 m. The garden area = 10 × 16 = 160 m². Let x metres be the width of the walk. The outer rectangle PQRS (garden plus walk) has dimensions:

Length of PQRS = 10 + 2x metres

Breadth of PQRS = 16 + 2x metres

Area of walk = Area of PQRS - Area of garden ABCD

120 = [(10 + 2x)(16 + 2x)] - 160

120 = 160 + 20x + 32x + 4x² - 160

120 = 4x² + 52x

4x² + 52x - 120 = 0

x² + 13x - 30 = 0

Factoring: x² + 15x - 2x - 30 = 0

x(x + 15) - 2(x + 15) = 0

(x - 2)(x + 15) = 0

This gives x = 2 or x = -15. Since width must be positive, x = 2 metres. Therefore, the equation is x² + 13x - 30 = 0 and the width of the walk is 2 metres.
In simple words: The walk's area is the difference between the outer rectangle (including walk) and the inner garden. We set up an equation, simplify it, and solve by factoring.

Exam Tip: Reject negative solutions—they are not physically meaningful for a width measurement. Always verify your answer by substituting back into the original walk area formula.

Question 15. A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface area of the four walls.
Answer: Given: Length = 6 m, breadth = 4.8 m, height = 3.5 m. The formula for the inner surface area of four walls is:

Inner surface area = 2(l + b) × h

Substituting the values:

= 2(6 + 4.8) × 3.5 = 2(10.8) × 3.5 = 21.6 × 3.5 = 75.6 m²

Therefore, the inner surface area of the four walls is 75.6 m².
In simple words: The four walls form a rectangle when unfolded. Their total area depends on the perimeter of the base and the height. Multiply the base perimeter by the height to get the surface area.

Exam Tip: This formula applies only to the four walls—ceiling and floor are not included. Use 2(l + b) for the perimeter, then multiply by height.

Question 16. A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.
Answer: Let ABCD be the rectangular plot with length 41 m and breadth 22.5 m. The boundary wall is constructed 1.5 m away from the plot on all sides, so the inner dimensions of the wall (the dimensions bordering the plot) are:

Length of inner face = 41 + 2(1.5) = 41 + 3 = 44 metres

Breadth of inner face = 22.5 + 2(1.5) = 22.5 + 3 = 25.5 metres

The height of the wall = 2 metres. Using the formula for perimeter of a rectangle:

Perimeter = 2(length + breadth) = 2(44 + 25.5) = 2(69.5) = 139 metres

Inner surface area of boundary wall = Perimeter × Height = 139 × 2 = 278 m²

Therefore, the inner surface area of the boundary wall is 278 m².
In simple words: The wall's inner surface touches the space around the plot. We find the inner perimeter (which includes the extra 1.5 m distance on each side), then multiply by the wall's height.

Exam Tip: The wall is built away from the plot, so the inner dimensions are larger than the plot dimensions by twice the distance (1.5 m on each opposite side). Visualize the layout carefully before calculating.

Question 17(a). Find the perimeter and area of the figure (i) given below in which all corners are right angles.
Answer: Looking at the figure, the shape is divided into two rectangles. The first rectangle ABFG has dimensions: length BF = BC + CF = 4 + 1.5 = 5.5 m, and width AB = 2 m. So its area is 5.5 × 2 = 11 m². The second rectangle CDEF has dimensions: length CD = 4 m and width DE = 1.5 m, giving an area of 4 × 1.5 = 6 m². Adding both areas: 11 + 6 = 17 m².

For the perimeter, we add all the outer sides. Since opposite sides of rectangles are equal, we have: AB + BC + CD + DE + EF + FG + AG = 2 + 4 + 4 + 1.5 + 4 + 2 + 5.5 = 23 m.
In simple words: Split the shape into two rectangles, find each area, and add them together. For the perimeter, just add up every edge you see on the outside.

Exam Tip: Always label all corners clearly and divide complex shapes into simpler rectangles - this reduces calculation errors and makes checking easier.

Question 17(b). Find the perimeter and area of the figure (ii) given below in which all corners are right angles.
Answer: The shape breaks down into three rectangles. Rectangle ABIJ has length AB = 3 m and width BI = BC + CI = 5 + 2 = 7 m, so area = 3 × 7 = 21 m². Rectangle EFGH has length EF = 2 m and width FG = 7 m, giving area = 2 × 7 = 14 m². Rectangle CDHI has length CD = 8 m and width DH = 2 m, so area = 8 × 2 = 16 m². Total area = 21 + 14 + 16 = 51 m².

For perimeter, we add all outer edges: AB + BC + CD + DE + EF + FG + GH + HI + IJ + JA = 3 + 5 + 8 + 5 + 2 + 7 + 2 + 8 + 3 + 7 = 50 m.
In simple words: Break the shape into three rectangles, work out each one's area, and add them. The perimeter is the sum of every side on the outside edge.

Exam Tip: Draw the figure with all corner points labeled (A, B, C, etc.) before calculating - this prevents mixing up lengths and helps you see which rectangles to add.

Question 17(c). Find the area and perimeter of the figure (iii) given below in which all corners are right angles and all measurements are in cm.
Answer: This complex shape splits into four rectangles. Rectangle BCDE has dimensions 5 × 2 = 10 cm². Rectangle FGHI has dimensions 3 × 2 = 6 cm². Rectangle JKLM has dimensions 5 × 2 = 10 cm². Rectangle ABMN has length AB = AC - BC = 7 - 5 = 2 cm and width AN = 12 cm, so area = 2 × 12 = 24 cm². Total area = 10 + 6 + 10 + 24 = 50 cm².

For the perimeter, the vertical distance from C to L equals the vertical distance from A to N. Accounting for this, the perimeter becomes: AC + DE + FG + HI + JK + LN + NA + CL = 7 + 5 + 3 + 3 + 5 + 7 + 12 + 12 = 54 cm.
In simple words: Divide the shape into four separate rectangles, find the area of each, then add all areas together. For the perimeter, go around the whole outside and add every side.

Exam Tip: When the figure has indentations, carefully identify which sides form the outer boundary - tracing the perimeter with your finger often prevents mistakes.

Question 18. The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one-third that of rectangle.
Answer: First, calculate the rectangle's area: area = length × breadth = 12 × 9 = 108 cm². The triangle's area equals one-third of this: triangle area = \( \frac{1}{3} \) × 108 = 36 cm².

Using the triangle area formula with base = 9 cm and height = h cm:
\( 36 = \frac{1}{2} \times 9 \times h \)

\( \implies 36 \times 2 = 9 \times h \)

\( \implies h = \frac{72}{9} = 8 \text{ cm} \)
In simple words: Find the rectangle's area first. Then divide it by three to get the triangle's area. Finally, use the triangle formula to work backwards and find the height.

Exam Tip: Always write the triangle area formula clearly and substitute numbers step-by-step - showing all working earns marks even if the final answer has an arithmetic slip.

Question 19. The area of a square plot is 484 m². Find the length of its one side and the length of its one diagonal.
Answer: Let the side length be x metres. Since area of square = side²:

\( 484 = x^2 \)

\( \implies x = \sqrt{484} = 22 \text{ m} \)

To find the diagonal, apply Pythagoras' theorem to the right triangle formed by two adjacent sides and the diagonal. If the diagonal is AC, then:

\( AC^2 = 22^2 + 22^2 = 484 + 484 = 968 \)

\( \implies AC = \sqrt{968} = 22\sqrt{2} \approx 22 \times 1.414 = 31.11 \text{ m} \)
In simple words: Take the square root of the area to get the side. Then use Pythagoras to find the diagonal by treating it as the hypotenuse of a right triangle made from two sides.

Exam Tip: Always simplify surds where possible (e.g., \( \sqrt{968} = 22\sqrt{2} \)) before approximating - examiners award marks for showing the exact form.

Question 20. A square has the perimeter 56 m. Find its area and the length of one diagonal correct up to two decimal places.
Answer: Starting with the perimeter formula: perimeter = 4 × side, so

\( 56 = 4x \implies x = 14 \text{ m} \)

The area of the square is:

\( \text{Area} = 14^2 = 196 \text{ m}^2 \)

To find the diagonal, use Pythagoras' theorem with two sides of 14 m each:

\( AC^2 = 14^2 + 14^2 = 196 + 196 = 392 \)

\( \implies AC = \sqrt{392} = 14\sqrt{2} \approx 14 \times 1.414 = 19.80 \text{ m} \)
In simple words: Divide the perimeter by 4 to get one side. Square that number to get the area. For the diagonal, use Pythagoras with both sides being the same length.

Exam Tip: Round to exactly two decimal places as requested - 19.796... becomes 19.80, not 19.8 - precision matters in marking schemes.

Question 21. A wire when bent in the form of an equilateral triangle encloses an area of 36√3 cm². Find the area enclosed by the same wire when bent to form: (i) a square, and (ii) a rectangle whose length is 2 cm more than its width.
Answer:

First, find the side of the equilateral triangle. Using the area formula:

\( \text{Area} = \frac{\sqrt{3}}{4} (\text{side})^2 \)

\( \implies \frac{\sqrt{3}}{4} (\text{side})^2 = 36\sqrt{3} \)

\( \implies (\text{side})^2 = \frac{36\sqrt{3} \times 4}{\sqrt{3}} = 144 \)

\( \implies \text{side} = 12 \text{ cm} \)

The perimeter of this triangle = 3 × 12 = 36 cm. This same wire is now reformed into different shapes.

(i) When bent into a square:

Perimeter of square = 36 cm, so side = 36 ÷ 4 = 9 cm.

Area of square = 9 × 9 = 81 cm².

(ii) When bent into a rectangle:

Perimeter of rectangle = 36 cm. Given that length is 2 cm more than width:

Let width = x cm, then length = (x + 2) cm.

Perimeter = 2(length + width):

\( 36 = 2[(x + 2) + x] \)

\( \implies 36 = 2[2x + 2] \)

\( \implies 36 = 4x + 4 \)

\( \implies 4x = 32 \implies x = 8 \text{ cm} \)

So width = 8 cm and length = 10 cm.

Area of rectangle = 10 × 8 = 80 cm².
In simple words: First work out the triangle's side using its area. This tells you the wire's length (the perimeter). Then bend that same wire into new shapes and calculate their areas using their new dimensions.

Exam Tip: Always show that perimeter stays constant when the wire is reformed - this is the key insight that links all three parts of the question together.

Question 22. Two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.
Answer: Consider the parallelogram ABCD where AB = 15 cm and BC = 10 cm. The distance between the longer sides is DM = 8 cm.

Using the formula for the area of a parallelogram:
Area = base × height = AB × DM = 15 × 8 = 120 cm².

Now, to find the distance between the shorter sides, let DN be that distance. Since the area remains the same regardless of which base and height we use:
Area = BC × DN
120 = 10 × DN
DN = 120 ÷ 10 = 12 cm

Therefore, the area of the parallelogram is 120 cm² and the distance between the shorter sides is 12 cm.
In simple words: A parallelogram's area stays the same no matter which side you call the base. Find it once using one base and height, then use that same area to find the other height.

Exam Tip: Remember that both methods of calculating area (base × height using different pairs) must give the same result - use this as a check on your work.

Question 23. ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.
Answer: To find the area, we first work with triangle ABC. Let the sides be: BC = a = 10 cm, AC = b = 16 cm, and AB = c = 12 cm.

Calculate the semi-perimeter:
s = (a + b + c) ÷ 2 = (10 + 16 + 12) ÷ 2 = 38 ÷ 2 = 19 cm

Apply Heron's formula for the area of the triangle:
Area of triangle = √[s(s - a)(s - b)(s - c)]
= √[19(19 - 10)(19 - 16)(19 - 12)]
= √[19 × 9 × 3 × 7]
= √3591
= 59.9 cm²

Since a diagonal splits a parallelogram into two triangles of equal area:
Area of parallelogram = 2 × Area of triangle ABC = 2 × 59.9 = 119.8 cm²

To find the distance between shorter sides (DM), use the area formula with base BC:
Area = BC × DM
119.8 = 10 × DM
DM = 119.8 ÷ 10 = 11.98 cm

The area of the parallelogram is 119.8 cm² and the distance between shorter sides is 11.98 cm.
In simple words: Split the parallelogram into two triangles using the diagonal, find one triangle's area with Heron's formula, then double it. Once you know the area, find the height using a different base.

Exam Tip: Always verify that Heron's formula gives a reasonable answer by checking whether (s - a), (s - b), and (s - c) are all positive.

Question 24. Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.
Answer: Let ABCD be a parallelogram with diagonals AC and BD meeting at point O. Given: AB = 12 cm and the perpendicular distance (height) DM = 6 cm.

First, find the area of the parallelogram:
Area of parallelogram ABCD = base × height = AB × DM = 12 × 6 = 72 cm²

Since the diagonals of a parallelogram bisect each other, O is the midpoint of BD. This means AO is a median of triangle ABD. A median divides a triangle into two triangles of equal area, so:
Area of triangle AOD = (1/2) × Area of triangle ABD ... (1)

Also, since the diagonal of a parallelogram divides it into two equal triangles:
Area of triangle ABD = (1/2) × Area of parallelogram ABCD

Substitute into equation (1):
Area of triangle AOD = (1/2) × (1/2) × Area of parallelogram ABCD
= (1/4) × 72
= 18 cm²

The area of triangle AOD is 18 cm².
In simple words: The diagonal cuts the parallelogram in half. The median (line from a corner through the midpoint of the opposite side) cuts that triangle in half again. So the final triangle is one quarter of the whole parallelogram.

Exam Tip: Key concept: diagonals bisect each other in a parallelogram, and a median always divides a triangle into two equal parts by area.

Question 25. ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.
Answer: Let ABCD be a parallelogram with diagonals AC and BD intersecting at O. Since the diagonals of a parallelogram bisect each other:
AO = 12 ÷ 2 = 6 cm and OB = 16 ÷ 2 = 8 cm

Look at triangle AOB with sides: AB = a = 10 cm, BO = b = 8 cm, and OA = c = 6 cm.

Calculate the semi-perimeter:
s = (a + b + c) ÷ 2 = (10 + 8 + 6) ÷ 2 = 24 ÷ 2 = 12 cm

Apply Heron's formula:
Area of triangle = √[s(s - a)(s - b)(s - c)]
= √[12(12 - 10)(12 - 8)(12 - 6)]
= √[12 × 2 × 4 × 6]
= √576
= 24 cm²

Since the diagonals of a parallelogram intersect at O, O is the midpoint of BD, making AO the median of triangle ABD. A median divides a triangle into two equal areas, so:
Area of triangle AOB = (1/2) × Area of triangle ABD ... (1)

Also, a diagonal divides the parallelogram into two equal triangles:
Area of triangle ABD = (1/2) × Area of parallelogram ABCD

From equation (1):
Area of triangle AOB = (1/2) × (1/2) × Area of parallelogram ABCD
24 = (1/4) × Area of parallelogram ABCD
Area of parallelogram ABCD = 24 × 4 = 96 cm²

The area of the parallelogram ABCD is 96 cm².
In simple words: One of the four triangles formed by the diagonals has area 24 cm². Since all four triangles are equal, the whole parallelogram has area four times that amount.

Exam Tip: In a parallelogram, diagonals create four triangles with equal areas - this relationship helps connect a single triangle's area to the whole shape's area.

Question 26. The area of a parallelogram is p cm² and its height is q cm. A second parallelogram has equal area but its base is r cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.
Answer: For the first parallelogram, use the area formula:
Area = base × height
p = base × q
base = p/q cm

The second parallelogram has a base that is r cm more than the first:
Base of 2nd parallelogram = (p/q + r) cm

The height of the second parallelogram is h cm. Since both parallelograms have the same area p cm²:
p = (p/q + r) × h

Solve for h:
p = (p + qr)/q × h
h = pq/(p + qr) cm

The height of the second parallelogram is h = pq/(p + qr) cm.
In simple words: If you make the base wider but keep the area the same, the height must get shorter. Rearrange the area formula to find how much shorter.

Exam Tip: When working with literal expressions, clearly isolate the required variable and simplify the fraction completely before stating the final answer.

Question 27. What is the area of a rhombus whose diagonals are 12 cm and 16 cm?
Answer: Use the formula for the area of a rhombus:
Area = (1/2) × d₁ × d₂

where d₁ and d₂ are the diagonals. Substituting d₁ = 12 cm and d₂ = 16 cm:
Area = (1/2) × 12 × 16 = (1/2) × 192 = 96 cm²

The area of the rhombus is 96 cm².
In simple words: The area of a rhombus is half the product of its two diagonals. Multiply the diagonals, then divide by 2.

Exam Tip: The rhombus area formula uses diagonals, not sides - this is different from most quadrilaterals. Make sure to use the correct lengths.

Question 28. The area of a rhombus is 98 cm². If one of its diagonal is 14 cm, what is the length of the other diagonal?
Answer: Use the area formula for a rhombus and rearrange to solve for the unknown diagonal:
Area = (1/2) × d₁ × d₂
98 = (1/2) × 14 × d₂
98 = 7 × d₂
d₂ = 98 ÷ 7 = 14 cm

The length of the other diagonal is 14 cm.
In simple words: Work backwards from the area formula. If you know the area and one diagonal, divide the area by half the known diagonal to find the other diagonal.

Exam Tip: In this case, both diagonals happen to be equal (14 cm), which means the rhombus is actually a square - a useful observation to verify your answer.

Question 29. The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.
Answer: First, find the length of each side. Since a rhombus has four equal sides:
Perimeter = 4 × side
45 = 4 × side
side = 45/4 = 11.25 cm

Now use the formula for the area of a rhombus (treating it like a parallelogram with a base and height):
Area = base × height
Area = 11.25 × 8 = 90 cm²

The area of the rhombus is 90 cm².
In simple words: Find each side by dividing the perimeter by 4. Then multiply that side length (the base) by the height to get the area.

Exam Tip: A rhombus can be thought of as a parallelogram, so the base × height formula applies. Use the perimeter to find the base length if it is not directly given.

Question 30. PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.
Answer: From the diagram, PQ is the base of rhombus PQRS and SM is the height (perpendicular distance).

Use the formula for the area of a rhombus:
Area = base × height
Area = PQ × SM
Area = 3 × 2.5
Area = 7.5 cm²

The area of rhombus PQRS is 7.5 cm².
In simple words: Multiply the side length by the perpendicular distance between two parallel sides to get the area.

Exam Tip: The height must be perpendicular to the base - it is not the same as the side length unless the rhombus is a square.

Question 31. If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.
Answer: Let ABCD be a rhombus where the diagonals AC = 8 cm and BD = 6 cm intersect at point O.

Since the diagonals of a rhombus bisect each other at right angles:
AO = 8 ÷ 2 = 4 cm and BO = 6 ÷ 2 = 3 cm

In the right-angled triangle AOB, apply the Pythagorean theorem to find the side length AB:
AB² = AO² + BO²
AB² = 4² + 3²
AB² = 16 + 9 = 25
AB = √25 = 5 cm

Since all sides of a rhombus are equal:
Perimeter = 4 × side = 4 × 5 = 20 cm

The perimeter of the rhombus is 20 cm.
In simple words: The diagonals split into halves at the center and meet at right angles. Use the Pythagorean theorem on one of the right triangles to find a side, then multiply by 4.

Exam Tip: The 3-4-5 Pythagorean triple appears here - recognizing common Pythagorean triples speeds up calculations and helps verify your answer.

Question 32. If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate (i) the length of the other diagonal, and (ii) the area of the rhombus.
Answer: (i) Let ABCD be a rhombus with diagonals AC and BD, where each side = 5 cm and one diagonal AC = 8 cm.

Since diagonals of a rhombus bisect each other at right angles:
AO = 8 ÷ 2 = 4 cm

In the right-angled triangle AOB, use the Pythagorean theorem:
AB² = AO² + BO²
5² = 4² + BO²
25 = 16 + BO²
BO² = 9
BO = 3 cm

Since the diagonals bisect each other:
BD = 2 × BO = 2 × 3 = 6 cm

The length of the other diagonal is 6 cm.

(ii) Now find the area using both diagonals:
Area = (1/2) × d₁ × d₂
Area = (1/2) × 8 × 6
Area = (1/2) × 48
Area = 24 cm²

The area of the rhombus is 24 cm².
In simple words: The diagonals split into halves and form right angles. Use Pythagorean theorem to find half of the unknown diagonal, then double it. Multiply both full diagonals and divide by 2 to get area.

Exam Tip: When one diagonal and all sides are known, the Pythagorean theorem in one right triangle gives the other diagonal's half-length - double this to get the full diagonal.

Question 33(a). The figure (i) given below is a trapezium. Find the length of BC and the area of the trapezium. Assume AB = 5 cm, AD = 4 cm, CD = 8 cm.
Answer: Start by drawing BN perpendicular to CD. This creates a rectangle BADN. In a rectangle, opposite sides are equal, so BN = AD = 4 cm and ND = BA = 5 cm. From the figure, CN = CD - ND = 8 - 5 = 3 cm. In the right-angled triangle BCN, apply Pythagoras' theorem: BC² = BN² + CN² = 4² + 3² = 16 + 9 = 25, giving BC = 5 cm. Using the trapezium formula, Area = (1/2) × (sum of parallel sides) × height = (1/2) × (5 + 8) × 4 = 26 cm².
In simple words: Draw a line from B at a right angle to the base. This helps you make a rectangle and a right triangle. Use the Pythagorean rule to find BC, then use the trapezium area formula.

Exam Tip: Always construct perpendiculars from the ends of the non-parallel sides to simplify the problem into rectangles and right triangles.

Question 33(b). The figure (ii) given below is a trapezium. Find (i) AB (ii) area of trapezium ABCD.
Answer: (i) Create a perpendicular from C to AD, parallel to AB, meeting AD at point M. Since ABCM is a rectangle, opposite sides must be equal: AM = CB = 2 units and CM = AB. From the figure, MD = AD - AM = 8 - 2 = 6 units. In right-angled triangle MDC, use Pythagoras' theorem: CD² = MD² + CM², so 10² = 6² + CM², giving 100 = 36 + CM², thus CM² = 64 and CM = 8 units. Since ABCM is a rectangle, AB = CM = 8 units. (ii) Apply the trapezium area formula: Area = (1/2) × (sum of parallel sides) × distance between them = (1/2) × (AD + BC) × AB = (1/2) × (8 + 2) × 8 = 40 sq. units.
In simple words: Drop a perpendicular from C to find the height. This creates a rectangle and a right triangle where you can use the Pythagorean rule. Then calculate the area using the trapezium formula.

Exam Tip: When finding unknown sides, construct perpendiculars strategically to form right triangles where Pythagoras' theorem can be applied directly.

Question 33(c). The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m², calculate its depth.
Answer: Think of ABCD as the cross-section of the canal shaped like a trapezium. Draw a perpendicular AM from A to CD, which represents the depth of the canal. The area of the cross-section equals 16.8 m². Using the trapezium area formula: (1/2) × (sum of parallel sides) × depth = 16.8. Substituting the known values: (1/2) × (6 + 8) × AM = 16.8, which simplifies to (1/2) × 14 × AM = 16.8. Solving for AM: AM = (16.8 × 2) / 14 = 33.6 / 14 = 2.4 m.
In simple words: The canal's cross-section is a trapezium with two parallel widths of 6 m and 8 m. Use the area formula with the given area to find the depth as 2.4 meters.

Exam Tip: Always rearrange the trapezium area formula to isolate the unknown variable (height or depth) when one is missing but the area is given.

Question 34. The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.
Answer: Let ABCD be the trapezium where AB is parallel to DC. Let E and F be the mid-points of sides AD and BC respectively, with EF = 18 cm. The perpendicular distance between the parallel sides is the height = 12 cm. A key property states: the sum of the lengths of two parallel sides equals 2 times the distance between the mid-points of the non-parallel sides. Therefore, AB + CD = 2 × EF = 2 × 18 = 36 cm. Using the area formula for a trapezium: Area = (1/2) × (sum of parallel sides) × height = (1/2) × 36 × 12 = 18 × 12 = 216 cm².
In simple words: The height is 12 cm and the distance between mid-points tells us the sum of the two parallel sides is 36 cm. Multiply these with the formula to get the area.

Exam Tip: Remember the mid-point property: the sum of parallel sides equals twice the distance between mid-points of non-parallel sides - this is a time-saving shortcut on exams.

Question 35. The area of a trapezium is 540 cm². If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Answer: Given: area = 540 cm², ratio of parallel sides = 7 : 5, and height = 18 cm. Let the parallel sides be 7x and 5x cm. Substitute into the area formula: 540 = (1/2) × (7x + 5x) × 18. Simplify: 540 = (1/2) × 12x × 18 = 6x × 18 = 108x. Solve: x = 540 / 108 = 5 cm. Therefore, the two parallel sides measure 7x = 7 × 5 = 35 cm and 5x = 5 × 5 = 25 cm.
In simple words: Use the ratio to write the sides as 7x and 5x. Put these into the area formula with the height to find x, then multiply by 7 and 5 to get each side.

Exam Tip: When sides are in a given ratio, always express them as multiples of a variable (like 7x and 5x) before substituting into the formula.

Question 36. The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm², find the perimeter.
Answer: Since ABCD is an isosceles trapezium, the non-parallel sides are equal: BC = AD. The parallel sides are in the ratio 2 : 3, so let CD = 2a and AB = 3a. From the area formula: 60 = (1/2) × (3a + 2a) × 4, which gives 60 = 2 × 5a = 10a, so a = 6 cm. This means AB = 18 cm and CD = 12 cm. Construct perpendiculars DN and CM from D and C to AB respectively. In the isosceles trapezium, triangles ADN and BCM are congruent (by RHS axiom), so AN = MB. Since DNMC is a rectangle, NM = DC = 12 cm. From AN + NM + MB = 18 and AN = MB, we get 2AN + 12 = 18, so AN = 3 cm. Using Pythagoras' theorem in right triangle AND: AD² = AN² + DN² = 3² + 4² = 9 + 16 = 25, giving AD = 5 cm. The perimeter = AB + BC + CD + DA = 18 + 5 + 12 + 5 = 40 cm.
In simple words: Set up sides as 2a and 3a, then use the area to find a = 6. Drop perpendiculars to create right triangles, use the Pythagorean rule to find the slant sides, then add all four sides.

Exam Tip: In isosceles trapeziums, the two non-parallel sides are always equal - use this to set up equal segments when perpendiculars are drawn.

Question 37. The area of a parallelogram is 98 cm². If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.
Answer: Let the base = x cm and the corresponding altitude = x/2 cm. Using the parallelogram area formula: Area = base × altitude, so 98 = x × (x/2) = x²/2. Multiply both sides by 2: x² = 196. Therefore, x = 14 cm. The base is 14 cm and the altitude is 14/2 = 7 cm. To verify: Area = 14 × 7 = 98 cm², which confirms our answer.
In simple words: The altitude is half the base, so if the base is x, the height is x/2. Put these in the area formula and solve for x.

Exam Tip: Always set up a relationship equation first when one measurement depends on another, then substitute into the relevant formula to solve.

Question 38. The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Answer: Let the breadth of the rectangular garden be x meters. Then the length is (x + 12) meters. The garden's area is x(x + 12) m². The perimeter is 2[(x + 12) + x] = (4x + 24) meters. According to the problem, the area equals 4 times the perimeter:

\( x(x + 12) = 4 \times (4x + 24) \)

\( x^2 + 12x = 16x + 96 \)

\( x^2 - 4x - 96 = 0 \)

\( x^2 - 12x + 8x - 96 = 0 \)

\( x(x - 12) + 8(x - 12) = 0 \)

\( (x + 8)(x - 12) = 0 \)

\( x = -8 \text{ or } x = 12 \)

Since breadth cannot be negative, x = 12 m. Therefore, breadth = 12 m and length = 24 m.
In simple words: Set up an equation where the area (length times breadth) equals 4 times the perimeter. Solve the quadratic equation to find that the breadth is 12 m and the length is 24 m.

Exam Tip: Always reject negative dimensions and verify your answer by checking that the area equals 4 times the perimeter using the calculated values.

Question 39. If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.
Answer: Let ABCD be a rectangular plot with length x m and breadth y m. From the perimeter formula:

\( 68 = 2(x + y) \)

\( x + y = 34 \)

\( x = 34 - y \text{ ... (1)} \)

In right triangle ABC, applying the Pythagorean theorem where the diagonal AC = 26 m:

\( AC^2 = AB^2 + BC^2 \)

\( 26^2 = x^2 + y^2 \)

\( x^2 + y^2 = 676 \)

Substituting equation (1):

\( (34 - y)^2 + y^2 = 676 \)

\( 1156 - 68y + y^2 + y^2 = 676 \)

\( 2y^2 - 68y + 480 = 0 \)

\( y^2 - 34y + 240 = 0 \)

\( y^2 - 24y - 10y + 240 = 0 \)

\( y(y - 24) - 10(y - 24) = 0 \)

\( (y - 10)(y - 24) = 0 \)

\( y = 10 \text{ m or } y = 24 \text{ m} \)

Substituting back into equation (1): when y = 10 m, x = 24 m; when y = 24 m, x = 10 m. In both cases, the area = 24 × 10 = 240 m².
In simple words: Use the perimeter to find a relationship between length and breadth, then use the Pythagorean theorem with the diagonal to solve for both dimensions. The area turns out to be 240 m² either way.

Exam Tip: Verify that both dimensions produce the same area to confirm your solution is correct.

Question 40. A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than a side of a square. Find the perimeter of the square.
Answer: Let the side of the square be x cm. Then the rectangle's length is (x + 12) cm and width is (x + 8) cm. Given that the rectangle's area is twice the square's area:

\( (x + 12)(x + 8) = 2x^2 \)

\( x^2 + 8x + 12x + 96 = 2x^2 \)

\( x^2 + 20x + 96 = 2x^2 \)

\( x^2 - 20x - 96 = 0 \)

\( x^2 - 24x + 4x - 96 = 0 \)

\( x(x - 24) + 4(x - 24) = 0 \)

\( (x + 4)(x - 24) = 0 \)

\( x = -4 \text{ or } x = 24 \text{ cm} \)

Since the side of a square cannot be negative, x = 24 cm. Therefore, the perimeter of the square = 4 × 24 = 96 cm.
In simple words: Create an equation showing that the rectangle's area is twice the square's area, then solve the resulting quadratic to find the square's side is 24 cm. The perimeter is 4 times this side.

Exam Tip: Always discard negative solutions for geometric dimensions and show the perimeter calculation as the final step.

Question 41. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm² less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.
Answer: Given the square's perimeter is 48 cm, its side length is 48 ÷ 4 = 12 cm. The square's area is 12² = 144 cm². Therefore, the rectangle's area is 144 - 4 = 140 cm². Let the breadth of the rectangle be x cm, so the length is (x + 4) cm. The area of the rectangle is:

\( x(x + 4) = 140 \)

\( x^2 + 4x - 140 = 0 \)

\( x^2 + 14x - 10x - 140 = 0 \)

\( x(x + 14) - 10(x + 14) = 0 \)

\( (x + 14)(x - 10) = 0 \)

\( x = -14 \text{ or } x = 10 \)

Since breadth cannot be negative, x = 10 cm. Therefore, breadth = 10 cm and length = 14 cm. The perimeter of the rectangle is 2(14 + 10) = 2 × 24 = 48 cm.
In simple words: Calculate the square's side and area first, then find the rectangle's area. Set up an equation with the breadth as the variable and solve to get length = 14 cm and breadth = 10 cm.

Exam Tip: Always calculate the square's dimensions and area before working on the rectangle, and verify your answer by checking the area difference.

Question 42. In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.
Answer: In right triangle AEB, let AE = BE = x cm. By the Pythagorean theorem:

\( AB^2 = AE^2 + EB^2 \)

\( 10^2 = x^2 + x^2 \)

\( 2x^2 = 100 \)

\( x^2 = 50 \)

\( x = 5\sqrt{2} \text{ cm} \)

Area of right triangle AEB = \( \frac{1}{2} \times x \times x = \frac{1}{2} \times 50 = 25 \text{ cm}^2 \)

In right triangle DCG, let DG = GC = y cm. Similarly:

\( 10^2 = y^2 + y^2 \)

\( y = 5\sqrt{2} \text{ cm} \)

Area of right triangle DCG = 25 cm². For equilateral triangles HAD and BFC with side 8 cm:

Area of each = \( \frac{\sqrt{3}}{4} \times 8^2 = 16\sqrt{3} \text{ cm}^2 \)

Area of rectangle ABCD = 10 × 8 = 80 cm². Total shaded area = 25 + 25 + 16√3 + 16√3 + 80 = \( 130 + 32\sqrt{3} \text{ cm}^2 \). Perimeter of figure = AE + EB + BF + FC + CG + GD + DH + HA = \( 5\sqrt{2} + 5\sqrt{2} + 8 + 8 + 5\sqrt{2} + 5\sqrt{2} + 8 + 8 = 20\sqrt{2} + 32 \text{ cm} \)
In simple words: Break the figure into simpler shapes - two right triangles, two equilateral triangles, and one rectangle - calculate each area and add them. For the perimeter, identify all the outer edges and sum their lengths.

Exam Tip: Label all calculated side lengths clearly (like the values of x and y) and show how each area formula is applied to its specific shape.

Question 43(a). Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimeters.
Answer: In right triangle ECF, applying the Pythagorean theorem:

\( EF^2 = EC^2 + CF^2 \)

\( 5^2 = EC^2 + 3^2 \)

\( EC^2 = 25 - 9 = 16 \)

\( EC = 4 \text{ cm} \)

Since DEFG is an isosceles trapezium, GD = EF = 5 cm. Since BDEC is a rectangle, BD = EC = 4 cm and BC = DE = 6 cm. In right triangle DBG:

\( GD^2 = BD^2 + GB^2 \)

\( 5^2 = 4^2 + GB^2 \)

\( GB^2 = 25 - 16 = 9 \)

\( GB = 3 \text{ cm} \)

In the trapezium, GF = GB + BC + CF = 3 + 6 + 3 = 12 cm. Area of trapezium DEFG = \( \frac{1}{2} \times (GF + DE) \times \text{height} = \frac{1}{2} \times (12 + 6) \times 4 = 36 \text{ cm}^2 \)

For equilateral triangle ABC with side 6 cm:

Area = \( \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \approx 15.59 \text{ cm}^2 \)

Total area = 36 + 15.59 = 51.59 cm².
In simple words: Use the Pythagorean theorem to find the missing dimensions of the right triangles, then use these to find the height of the trapezium. Calculate the trapezium and triangle areas separately, then add them together.

Exam Tip: Identify all the geometric shapes that make up the composite figure and use the appropriate area formulas for each shape individually before summing them.

Question 43(b). Find the area enclosed by the figure (ii) given below. All measurements are in centimeters.
Answer: Looking at the figure, it consists of a vertical rectangle and two triangular regions projecting outward on the right side. The vertical rectangle has width 2 cm and height (2 + 2 + 2 + 2) = 8 cm, giving an area of 2 × 8 = 16 cm². The upper triangular region has a base of 2 cm and a height that can be determined from the geometry. Since the total width extends to 6 cm and the rectangle is 2 cm wide, the maximum extension is 4 cm. For the upper triangle, the height is 2 cm, so its area is \( \frac{1}{2} \times 4 \times 2 = 4 \text{ cm}^2 \). Similarly, the lower triangle has area \( \frac{1}{2} \times 4 \times 2 = 4 \text{ cm}^2 \). Total area = 16 + 4 + 4 = 24 cm².
In simple words: Divide the composite figure into simpler pieces - one rectangle in the middle and two triangles on the sides. Calculate the area of each piece using basic formulas and add them together.

Exam Tip: When dealing with composite figures, always divide them into recognizable shapes and label the dimensions clearly so the calculation steps are easy to follow.

Question 43(c). In the figure (iii) given below, from a 24 cm × 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.
Answer: Looking at the figure, the cardboard is split into four sections: two rectangles (I and II) and two parallelograms (III and IV).

Rectangle (I) has length 24 cm and breadth 6 cm, so its area is 24 × 6 = 144 cm².

Rectangle (II) also has length 24 cm and breadth 6 cm, making its area 24 × 6 = 144 cm².

Parallelogram (III) has base 8 cm and height 6 cm, giving area 8 × 6 = 48 cm².

Parallelogram (IV) also has base 8 cm and height 6 cm, so its area is 8 × 6 = 48 cm².

The total area of the M-shaped figure is 144 + 144 + 48 + 48 = 384 cm².

The cardboard piece is 24 × 24 = 576 cm².

The remaining cardboard after cutting is 576 - 384 = 192 cm².
In simple words: Find the area of the whole square cardboard. Find the area of the M shape by breaking it into rectangles and parallelograms. Subtract the M shape from the whole square.

Exam Tip: Always break complex shapes into simpler known shapes like rectangles, triangles, and parallelograms. Double-check that all component areas add up correctly before subtracting from the total.

Question 44(a). The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.
Answer: The cross-section is made up of a trapezium and a rectangle positioned together.

For the trapezium: The two parallel sides are 0.6 m and 1.5 m. The perpendicular distance between them is 1.2 + 2.4 = 3.6 m.

Area of trapezium = \( \frac{1}{2} \times (0.6 + 1.5) \times 3.6 = \frac{1}{2} \times 2.1 \times 3.6 = 2.1 \times 1.8 = 3.78 \) m².

For the rectangle: Length is 2.4 m and breadth is 0.3 m.

Area of rectangle = 2.4 × 0.3 = 0.72 m².

Total cross-sectional area = 3.78 + 0.72 = 4.5 m².
In simple words: Split the shape into a trapezium and rectangle. Calculate each area using its own formula, then add them together to get the total area.

Exam Tip: When dealing with composite figures, identify the component shapes first and apply the correct formula for each. Ensure measurements are in the same units before performing calculations.

Question 44(b). The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.
Answer: The field is made up of four distinct regions: a right-angled triangle AXB, a trapezium XZCB, a right-angled triangle CZD, and triangle AED.

Triangle AXB: This is a right-angled triangle with base BX = 30 m and height AX = 12 m.

Area = \( \frac{1}{2} \times 30 \times 12 = 180 \) m².

Trapezium XZCB: The parallel sides are BX = 30 m and CZ = 25 m. The perpendicular distance is 15 m.

Area = \( \frac{1}{2} \times (30 + 25) \times 15 = \frac{1}{2} \times 55 \times 15 = 412.5 \) m².

Triangle CZD: This is a right-angled triangle with base CZ = 25 m and height ZD = 10 m.

Area = \( \frac{1}{2} \times 25 \times 10 = 125 \) m².

Triangle AED: Base AD = 37 m and height EY = 20 m (where Y is the perpendicular from E to AD).

Area = \( \frac{1}{2} \times 37 \times 20 = 370 \) m².

Total field area = 180 + 412.5 + 125 + 370 = 1087.5 m².
In simple words: Divide the field into triangles and a trapezium. Work out the area of each piece using base and height. Add all the pieces together.

Exam Tip: Always carefully identify which segments are bases and heights in composite figures. Use perpendicular distances, not slant heights, for area calculations.

Question 44(c). Calculate the area of the pentagon ABCDE shown in fig (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5 cm, DX = 9 cm and DX is perpendicular to EC and AB.
Answer: The pentagon is divided into three parts: two triangles (DEY and DYC) and one trapezium (ECBA).

For triangle DEY: Using the Pythagorean theorem in the right-angled triangle,

\( DE^2 = DY^2 + EY^2 \)

\( 5^2 = DY^2 + 4^2 \)

\( 25 = DY^2 + 16 \)

\( DY^2 = 9 \)

\( DY = 3 \) cm.

Area of triangle DEY = \( \frac{1}{2} \times 4 \times 3 = 6 \) cm².

For triangle DYC: Since DC = 5 cm and the right angle is at Y,

Area of triangle DYC = \( \frac{1}{2} \times 4 \times 3 = 6 \) cm².

For trapezium ECBA: XY = DX - DY = 9 - 3 = 6 cm. The parallel sides are EC = EY + CY = 4 + 4 = 8 cm and AB = AX + BX = 6 + 6 = 12 cm. The perpendicular distance is 6 cm.

Area of trapezium = \( \frac{1}{2} \times (8 + 12) \times 6 = \frac{1}{2} \times 20 \times 6 = 60 \) cm².

Total area of pentagon = 6 + 6 + 60 = 72 cm².
In simple words: Find the height DY using the Pythagorean theorem. Calculate the two triangle areas. Then calculate the trapezium area. Add all three areas.

Exam Tip: When perpendicularity is given, use it to set up right triangles and apply the Pythagorean theorem. Always verify that all component measurements match before summing.

Question 45. If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.
Answer: Let the length be l metres and breadth be b metres. The original area is lb m².

From the first condition: When both length and breadth increase by 1 m, the area grows by 21 m².

\( (l + 1)(b + 1) - lb = 21 \)

\( lb + l + b + 1 - lb = 21 \)

\( l + b = 20 \) ... (1)

From the second condition: When length increases by 1 m and breadth decreases by 1 m, the area shrinks by 5 m².

\( lb - (l + 1)(b - 1) = 5 \)

\( lb - (lb - l + b - 1) = 5 \)

\( l - b = 4 \) ... (2)

Adding equations (1) and (2):

\( 2l = 24 \)

\( l = 12 \) m.

Substituting into equation (2):

\( 12 - b = 4 \)

\( b = 8 \) m.

Perimeter = \( 2(l + b) = 2 \times 20 = 40 \) m.
In simple words: Set up two equations from the two conditions given. Solve them together to find length and breadth. Then use the perimeter formula.

Exam Tip: Always expand algebraic expressions fully before simplifying. Verify your answer by substituting the values back into the original conditions.

Question 46. A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer: Let the sides of the triangle be a = 26 cm, b = 28 cm, and c = 30 cm.

The semi-perimeter is:

\( s = \frac{a + b + c}{2} = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42 \) cm.

Using Heron's formula, the area of the triangle is:

\( \text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \)

\( = \sqrt{42 \times 16 \times 14 \times 12} \)

\( = \sqrt{112896} \)

\( = 336 \) cm².

Since the parallelogram has the same area and stands on base 28 cm:

\( \text{Area of parallelogram} = \text{base} \times \text{height} \)

\( 336 = 28 \times h \)

\( h = \frac{336}{28} = 12 \) cm.
In simple words: Use Heron's formula to find the triangle's area from its three sides. Since the parallelogram has the same area, divide that area by the base 28 to get the height.

Exam Tip: Heron's formula is essential when the triangle height is not directly given. Always compute the semi-perimeter first, then carefully substitute into the formula. Double-check the final division.

Question 47. A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.
Answer: Let the length be x cm.

Since the area is 105 cm², the breadth is \( \frac{105}{x} \) cm.

The perimeter of the rectangle is given as 44 cm.

Using the perimeter formula:

\( \text{Perimeter} = 2(\text{length} + \text{breadth}) \)

\( 44 = 2\left(x + \frac{105}{x}\right) \)

\( 22 = x + \frac{105}{x} \)

Multiplying both sides by x:

\( 22x = x^2 + 105 \)

\( x^2 - 22x + 105 = 0 \)

Using the quadratic formula or factoring:

\( (x - 7)(x - 15) = 0 \)

\( x = 7 \text{ or } x = 15 \)

If x = 7 cm (length), then breadth = \( \frac{105}{7} = 15 \) cm.

If x = 15 cm (length), then breadth = \( \frac{105}{15} = 7 \) cm.

Both solutions give a rectangle with dimensions 15 cm × 7 cm (length can be either dimension).
In simple words: Express breadth using area and length. Use the perimeter condition to form an equation. Solve the resulting quadratic equation to find length, then calculate breadth.

Exam Tip: When expressing one dimension in terms of another using area, always set up the perimeter equation carefully. Quadratic equations often have two solutions - both may be valid, but check which is more meaningful in context (here, both work, just swapping length and breadth).

Question 48. The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Answer: Let the length of the rectangle be x meters. We know the perimeter is 180 m. From the perimeter formula: \( 2(l + b) = 180 \)
\( \Rightarrow 2(x + b) = 180 \)
\( \Rightarrow x + b = 90 \)
\( \Rightarrow b = (90 - x) \) m.
Now, the area of the rectangle is given as 1800 m²:
\( \text{Area} = l \times b \)
\( \Rightarrow x(90 - x) = 1800 \)
\( \Rightarrow 90x - x^2 = 1800 \)
\( \Rightarrow x^2 - 90x + 1800 = 0 \)
\( \Rightarrow x^2 - 60x - 30x + 1800 = 0 \)
\( \Rightarrow x(x - 60) - 30(x - 60) = 0 \)
\( \Rightarrow (x - 30)(x - 60) = 0 \)
\( \Rightarrow x = 30 \text{ or } x = 60 \)
When x = 30, the breadth becomes 90 - 30 = 60 m. When x = 60, the breadth becomes 90 - 60 = 30 m. Therefore, the breadth is (90 - x) m, the equation is \( x(90 - x) = 1800 \), and the dimensions of the rectangle are 60 m by 30 m.
In simple words: Use the perimeter to find breadth in terms of length. Write an equation using area. Solve to get two possible values, giving dimensions of either 60 m × 30 m.

Exam Tip: Always express breadth using the perimeter first, then substitute into the area equation. Both solutions (x = 30 and x = 60) are valid since they represent the same rectangle with length and breadth swapped.

Exercise 15.3

Question 1. Find the length of the diameter of a circle whose circumference is 44 cm.
Answer: Let the radius be r cm. Using the circumference formula:
\( \text{Circumference} = 2\pi r \)
\( \Rightarrow 2\pi r = 44 \)
\( \Rightarrow 2 \times \frac{22}{7} \times r = 44 \)
\( \Rightarrow \frac{44r}{7} = 44 \)
\( \Rightarrow r = \frac{44 \times 7}{44} = 7 \) cm.
Since diameter = 2r:
\( \text{Diameter} = 2 \times 7 = 14 \) cm. Therefore, the diameter of the circle is 14 cm.
In simple words: The circumference tells us the distance around the circle. Divide the circumference by 2π to find the radius, then multiply by 2 to get the diameter.

Exam Tip: Remember that circumference \( = 2\pi r \) and diameter \( = 2r \). Always ensure you use the correct value of π (typically 22/7 or 3.14) as specified in the question.

Question 2. Find the radius and area of a circle if its circumference is 18π cm.
Answer: Let the radius be r cm. From the circumference formula:
\( \text{Circumference} = 2\pi r \)
\( \Rightarrow 2\pi r = 18\pi \)
\( \Rightarrow 2r = 18 \)
\( \Rightarrow r = 9 \) cm.
Now, the area of the circle:
\( \text{Area} = \pi r^2 = \frac{22}{7} \times 9^2 = \frac{22 \times 81}{7} = \frac{1782}{7} = 254\frac{4}{7} \) cm². Therefore, the radius is 9 cm and the area is \( 254\frac{4}{7} \) cm².
In simple words: When circumference is expressed in terms of π, divide by 2π to find the radius. Then square the radius and multiply by π to get the area.

Exam Tip: When the circumference includes π in the answer, simplify by cancelling π from both sides of the equation before solving. This makes the calculation much simpler.

Question 3. Find the perimeter of a semicircular plate of radius 3.85 cm.
Answer: The perimeter of a semicircle includes the curved part and the straight diameter. Using the formula:
\( \text{Perimeter of semicircular plate} = (\pi + 2)r \)
\( = \left(\frac{22}{7} + 2\right) \times 3.85 \)
\( = \frac{22 + 14}{7} \times 3.85 \)
\( = \frac{36}{7} \times 3.85 \)
\( = 36 \times 0.55 \)
\( = 19.8 \) cm. Therefore, the perimeter of the semicircular plate is 19.8 cm.
In simple words: A semicircle's perimeter is half the circle's circumference plus the straight line across. Add πr (half the full circumference) to 2r (the diameter).

Exam Tip: Do not forget to include the straight edge (diameter = 2r) when finding the perimeter of a semicircle. This is a common mistake.

Question 4. Find the radius and circumference of a circle whose area is 144π cm².
Answer: Let the radius be r cm. From the area formula:
\( \text{Area} = \pi r^2 \)
\( \Rightarrow 144\pi = \pi r^2 \)
\( \Rightarrow r^2 = 144 \)
\( \Rightarrow r = \sqrt{144} = 12 \) cm.
Now, the circumference:
\( \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} = 75\frac{3}{7} \) cm. Therefore, the radius is 12 cm and the circumference is \( 75\frac{3}{7} \) cm.
In simple words: When the area is given in terms of π, divide by π to find r². Take the square root to get the radius. Then use the circumference formula with this radius value.

Exam Tip: When area includes π, cancel it out immediately before taking the square root - this avoids unnecessary complexity in the calculation.

Question 5. A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
Answer: From the problem, each circular piece has a diameter of 0.5 cm. To fit circles efficiently on a rectangular sheet, we pack them in a grid where each circle is enclosed in a square. The side length of each square equals the diameter: 0.5 cm.
The number of such squares (and therefore discs) that fit:
\( \text{No. of squares} = \frac{\text{Area of sheet}}{\text{Area of a square}} = \frac{11 \times 2}{0.5 \times 0.5} = \frac{22}{0.25} = 88 \)
Therefore, 88 discs can be prepared from the sheet.
In simple words: Each disc needs a square space equal to its diameter. Divide the total sheet area by the area of one square to find how many fit.

Exam Tip: In packing problems, think of fitting each circle inside a square (side = diameter). This grid approach gives the maximum number of complete discs that can be cut without overlap.

Question 6. If the area of the semi-circular region is 77 cm², find its perimeter.
Answer: Let the radius be r cm. From the area of a semicircle:
\( \frac{\pi r^2}{2} = 77 \)
\( \Rightarrow \pi r^2 = 154 \)
\( \Rightarrow \frac{22}{7} \times r^2 = 154 \)
\( \Rightarrow r^2 = \frac{154 \times 7}{22} = 49 \)
\( \Rightarrow r = 7 \) cm.
Now, the perimeter of the semicircle:
\( \text{Perimeter} = (\pi + 2)r = \frac{22}{7} \times 7 + 2 \times 7 = 22 + 14 = 36 \) cm. Therefore, the perimeter of the semicircular region is 36 cm.
In simple words: Start by finding the radius using the area formula. Then add the curved part (πr) and the straight edge (2r) to get the perimeter.

Exam Tip: When given the area of a semicircle, always rearrange to find r² first. The perimeter formula combines half the full circumference plus the diameter.

Question 7(a). In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the area of shaded portion is 308 cm², calculate: (i) the length of AC and (ii) the circumference of the circle.
Answer: (i) Let r be the radius of the circle. Two perpendicular diameters divide a circle into four equal sections (quadrants), each with area \( \frac{1}{4}\pi r^2 \). Since two quadrants are shaded:
\( \text{Area of shaded region} = 2 \times \frac{1}{4}\pi r^2 = \frac{1}{2}\pi r^2 \)
\( \Rightarrow 308 = \frac{1}{2} \times \frac{22}{7} \times r^2 \)
\( \Rightarrow r^2 = \frac{308 \times 2 \times 7}{22} = \frac{308 \times 14}{22} = 196 \)
\( \Rightarrow r = 14 \) cm.
Since AC is a diameter:
\( \text{AC} = 2r = 2 \times 14 = 28 \) cm.
(ii) The circumference:
\( \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \) cm. Therefore, AC = 28 cm and the circumference is 88 cm.
In simple words: Two perpendicular diameters create four equal pieces. If two are shaded, use that area to find the radius. Then double it for diameter and apply the circumference formula.

Exam Tip: Always identify how many quadrants or sectors are shaded to set up the correct area equation. Remember that two perpendicular diameters create four equal parts.

Question 7(b). In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with center O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take π = 3.14)
Answer: Given AC = 16 cm, so the radius \( r = \frac{16}{2} = 8 \) cm. Two perpendicular diameters divide the circle into four equal quadrants.
Area of each quadrant:
\( \text{Area of each quadrant} = \frac{\pi r^2}{4} = \frac{3.14 \times 8^2}{4} = \frac{3.14 \times 64}{4} = 3.14 \times 16 = 50.24 \) cm².
Area of two shaded quadrants:
\( \text{Area of shaded region} = 2 \times 50.24 = 100.48 \) cm².
Perimeter of each quadrant includes a quarter-circle arc and two radii:
\( \text{Perimeter of each quadrant} = \frac{2\pi r}{4} + 2r = \frac{\pi r}{2} + 2r = \frac{3.14 \times 8}{2} + 2 \times 8 = 12.56 + 16 = 28.56 \) cm.
Perimeter of both quadrants:
\( \text{Perimeter of shaded region} = 2 \times 28.56 = 57.12 \) cm. Therefore, the area of the shaded region is 100.48 cm² and the perimeter is 57.12 cm.
In simple words: Each quadrant is one-quarter of the circle. Its perimeter has an arc (quarter of the circumference) and two straight radii. Add them up for one quadrant, then double for two quadrants.

Exam Tip: For quadrant perimeter, include both the curved arc and the two straight radii. Do not forget the radii - this is where students often lose marks.

Question 8. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.
Answer: First, find the total time:
\( \text{Time} = 1 \text{ minute } 28 \text{ seconds} = 60 + 28 = 88 \text{ seconds} \)
Distance the bucket travels:
\( \text{Distance} = \text{Speed} \times \text{Time} = 1.1 \times 88 = 96.8 \text{ m} \)
Radius of the wheel:
\( r = \frac{\text{Diameter}}{2} = \frac{77}{2} = 38.5 \text{ cm} \)
Circumference of the wheel:
\( \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 38.5 = 242 \text{ cm} = 2.42 \text{ m} \)
Number of revolutions:
\( \text{Number of revolutions} = \frac{\text{Distance travelled}}{\text{Circumference}} = \frac{96.8}{2.42} = 40 \)
Therefore, the wheel makes 40 complete revolutions in raising the bucket.
In simple words: Find how far the bucket travels using speed and time. One wheel rotation covers a distance equal to the circumference. Divide total distance by circumference to get the number of rotations.

Exam Tip: Always convert time and distance to consistent units. Once you have the circumference, simple division gives the number of revolutions needed.

Question 9. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer, correct to the nearest km.
Answer: Radius of the wheel:
\( r = \frac{\text{Diameter}}{2} = \frac{84}{2} = 42 \text{ cm} \)
Distance covered in one revolution:
\( \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 42 = 264 \text{ cm} \)
Distance covered in 5 revolutions:
\( \text{Distance} = 5 \times 264 = 1320 \text{ cm} \)
The wheel covers 1320 cm in 1 second. Converting to km/hr:
\( \text{Speed} = \frac{1320 \times 10^{-5}}{1/(60 \times 60)} = 1320 \times 60 \times 60 \times 10^{-5} = 4752000 \times 10^{-5} = 47.52 \approx 48 \text{ km/hr} \)
Therefore, the speed of the cart is approximately 48 km/hr.
In simple words: Find the circumference to know the distance per rotation. Multiply by the number of revolutions per second to get distance per second. Convert to km/hr using the conversion factor 3.6.

Exam Tip: To convert cm/sec to km/hr, multiply by 3.6 (or multiply by 60 × 60 and divide by 100000). Always round to the nearest km as asked in the question.

Question 10. The circumference of a circle is 123.2 cm. Calculate: (i) the radius of the circle in cm.
Answer: From the circumference formula:
\( \text{Circumference} = 2\pi r \)
\( \Rightarrow 123.2 = 2 \times \frac{22}{7} \times r \)
\( \Rightarrow 123.2 = \frac{44}{7} \times r \)
\( \Rightarrow r = \frac{123.2 \times 7}{44} = \frac{862.4}{44} = 19.6 \text{ cm} \)
Therefore, the radius of the circle is 19.6 cm.
In simple words: Use the circumference formula and solve for r by dividing the given circumference by 2π.

Exam Tip: When given circumference, always rearrange \( 2\pi r = C \) and solve for r directly. Check your answer by multiplying r by 2π to verify you get back the original circumference.

Question. (ii) Find the area of the circle in cm², correct to the nearest cm².
Answer: Using the formula Area = πr², we get:
\( \text{Area} = \frac{22}{7} \times (19.6)^2 = \frac{22}{7} \times 384.16 = 22 \times 54.88 = 1207.36 \approx 1207 \text{ cm}^2 \)
In simple words: Multiply pi by the square of the radius. The answer comes to 1207 square centimetres when you round it.

Exam Tip: Always round to the nearest whole number as instructed. Check that your radius value from part (i) is correct before squaring it.

Question. (iii) Find the effect on the area of the circle if the radius is doubled.
Answer: Let the original radius be r. The original area is πr². If we double the radius to 2r, the new area becomes π(2r)² = 4πr². Dividing the new area by the original area:
\( \frac{4\pi r^2}{\pi r^2} = 4 \)
The area becomes 4 times larger.
In simple words: When you double the radius, the area grows to four times what it was before.

Exam Tip: Remember that area depends on the radius squared, so doubling the radius multiplies the area by 2² = 4. This is a fundamental relationship to memorise.

Question 11(a). In the figure (i) given below, the area enclosed between the concentric circles is 770 cm². Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
Answer: Let the radius of the inner circle be r cm. The shaded region (the ring between the circles) equals the difference of the two circle areas:
\( 770 = \pi(21)^2 - \pi r^2 \)
\( 770 = 441\pi - \pi r^2 \)
\( 770 = \pi(441 - r^2) \)
\( 441 - r^2 = \frac{770}{\pi} = \frac{770}{\frac{22}{7}} = \frac{770 \times 7}{22} = 245 \)
\( r^2 = 441 - 245 = 196 \)
\( r = \sqrt{196} = 14 \text{ cm} \)
In simple words: Subtract the area of the smaller circle from the larger circle to get 770. This helps you find that the smaller radius is 14 cm.

Exam Tip: The shaded area between two concentric circles always equals the difference of their areas. Be careful to use the correct formula for π when substituting values.

Question 11(b). In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm². The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.
Answer: Let r = radius of the inner circle and R = radius of the outer circle. From the circumference of the inner circle:
\( 2\pi r = 88 \)
\( r = \frac{88}{2\pi} = \frac{44}{\frac{22}{7}} = \frac{44 \times 7}{22} = 14 \text{ cm} \)
The shaded area between the circles equals:
\( 346.5 = \pi R^2 - \pi(14)^2 \)
\( 346.5 = \pi(R^2 - 196) \)
\( R^2 - 196 = \frac{346.5}{\pi} = \frac{346.5}{\frac{22}{7}} = \frac{346.5 \times 7}{22} = 110.25 \)
\( R^2 = 110.25 + 196 = 306.25 \)
\( R = \sqrt{306.25} = 17.5 \text{ cm} \)
In simple words: Use the circumference formula to find the inner radius. Then use the area difference to find the outer radius, which is 17.5 cm.

Exam Tip: When given a circumference, always use 2πr to find the radius first. Then apply the area difference formula with your calculated inner radius.

Question 12. A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at Rs 200 per m².
Answer: First, find the radius of the circular plot from its circumference:
\( 2\pi R = 44 \)
\( 2 \times \frac{22}{7} \times R = 44 \)
\( R = \frac{44 \times 7}{44} = 7 \text{ m} \)
The outer radius (including the road) is 7 + 3.5 = 10.5 m. The area of the road is:
\( \text{Area of road} = \pi(10.5)^2 - \pi(7)^2 = 110.25\pi - 49\pi = 61.25\pi \)
\( = \frac{22}{7} \times 61.25 = 192.50 \text{ m}^2 \)
\( \text{Cost} = 192.50 \times 200 = \text{Rs } 38,500 \)
In simple words: Find both radii, calculate the ring-shaped area between them, then multiply by the cost per square metre.

Exam Tip: Don't forget to add the road width to the inner radius to get the outer radius. Always check your area calculation using the difference of two circles.

Question 13. The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumferences of the two circles.
Answer: Let R and r be the radii of the larger and smaller circles. From the given information:
\( 2R + 2r = 14 \implies R + r = 7 \quad \text{...(1)} \)
\( 2\pi R - 2\pi r = 8 \implies \pi(R - r) = 4 \)
\( \frac{22}{7}(R - r) = 4 \implies R - r = \frac{28}{22} \quad \text{...(2)} \)
Adding (1) and (2):
\( 2R = 7 + \frac{28}{22} = \frac{154 + 28}{22} = \frac{182}{22} \implies R = \frac{91}{22} \text{ cm} \)
From (2): \( r = \frac{91}{22} - \frac{28}{22} = \frac{63}{22} \text{ cm} \)
Circumference of larger circle: \( 2\pi R = 2 \times \frac{22}{7} \times \frac{91}{22} = 2 \times 13 = 26 \text{ cm} \)
Circumference of smaller circle: \( 2\pi r = 2 \times \frac{22}{7} \times \frac{63}{22} = 2 \times 9 = 18 \text{ cm} \)
In simple words: Write two equations from the two given facts. Solve them together to get both radii, then find the circumferences.

Exam Tip: Set up two separate equations carefully - one from diameters and one from circumferences. Use substitution or addition to solve the system.

Question 14. Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.
Answer: Let r be the radius of the resultant circle. The area of this circle equals the sum of the three given areas:
\( \pi r^2 = \pi(2)^2 + \pi(3)^2 + \pi(6)^2 \)
\( \pi r^2 = \pi(4 + 9 + 36) \)
\( r^2 = 49 \)
\( r = 7 \text{ cm} \)
Circumference: \( 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm} \)
In simple words: Add up the areas of all three circles, then use that total to find the radius of one big circle.

Exam Tip: This problem combines the areas of three separate circles into one. Remember that π cancels out when you divide both sides, simplifying the calculation.

Question 15. A copper wire when bent in the form of a square encloses an area of 121 cm². If the same wire is bent into the form of a circle, find the area of the circle.
Answer: From the square's area:
\( (\text{side})^2 = 121 \implies \text{side} = 11 \text{ cm} \)
Perimeter of square: \( 4 \times 11 = 44 \text{ cm} \)
This same wire length becomes the circumference of the circle:
\( 2\pi r = 44 \)
\( r = \frac{44}{2\pi} = \frac{44}{2 \times \frac{22}{7}} = \frac{44 \times 7}{44} = 7 \text{ cm} \)
Area of circle: \( \pi r^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2 \)
In simple words: The wire's length stays the same whether it forms a square or circle. Use this length to find the circle's radius, then its area.

Exam Tip: In these "same wire" problems, the perimeter/circumference is constant. This is the key relationship that connects the two shapes.

Question 16. A copper wire when bent in the form of an equilateral triangle has area \( 121\sqrt{3} \) cm². If the same wire is bent in the form of a circle, find the area enclosed by the wire.
Answer: For an equilateral triangle with side s:
\( \text{Area} = \frac{\sqrt{3}}{4}(s)^2 \)
Given area = \( 121\sqrt{3} \):
\( \frac{\sqrt{3}}{4}(s)^2 = 121\sqrt{3} \)
\( (s)^2 = 484 \implies s = 22 \text{ cm} \)
Perimeter of triangle: \( 3 \times 22 = 66 \text{ cm} \)
This becomes the circle's circumference:
\( 2\pi r = 66 \)
\( r = \frac{66 \times 7}{2 \times 22} = \frac{21}{2} \text{ cm} \)
Area of circle: \( \pi r^2 = \frac{22}{7} \times \left(\frac{21}{2}\right)^2 = \frac{22}{7} \times \frac{441}{4} = 346.5 \text{ cm}^2 \)
In simple words: Use the equilateral triangle area formula to find the side. The perimeter becomes the circle's circumference, which leads to its area.

Exam Tip: Memorise the area formula for an equilateral triangle: \( \frac{\sqrt{3}}{4}s^2 \). Remember to simplify fractions carefully when dealing with radius values.

Question 17(a). Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.
Answer: Radius of the smaller circle with diameter 7 cm:
\( r_{\text{small}} = \frac{7}{2} = 3.5 \text{ cm} \)
Area of larger circle = 16 times the area of smaller circle:
\( \pi R^2 = 16 \times \pi(3.5)^2 \)
\( R^2 = 16 \times 12.25 = 196 \)
\( R = 14 \text{ cm} \)
Circumference: \( 2\pi R = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \)
In simple words: Find the area of the small circle first. Then find what radius gives an area 16 times bigger, and calculate the circumference.

Exam Tip: When areas are in a ratio of 16:1, the radii are in a ratio of 4:1 (since area depends on r²). Use this relationship to speed up your calculation.

Question 17(b). In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 3.14)
Answer: From the figure, three circles of radius 3 cm are arranged inside a rectangle. The rectangle's width equals the diameter of one circle (6 cm), and its length equals the diameter of three circles side by side (18 cm).
Area of rectangle: \( 18 \times 6 = 108 \text{ cm}^2 \)
Total area of three circles: \( 3 \times \pi r^2 = 3 \times 3.14 \times 9 = 84.78 \text{ cm}^2 \)
Unshaded area: \( 108 - 84.78 = 23.22 \text{ cm}^2 \)
In simple words: Calculate the whole rectangle's area, then subtract the total area of all three circles that sit inside it.

Exam Tip: Always identify the dimensions of the rectangle carefully from the figure. Use the radius value shown to calculate the circle areas accurately.

Question 18. In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircle are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = 22/7.
Answer: The square ABCD has an area of (21)² = 441 cm². When the diagonals of a square intersect, they divide it into four triangles with equal areas. Each triangle has an area of 441/4 = 110.25 cm². The sum of two opposite triangles (△AOD and △BOC) equals 110.25 + 110.25 = 220.50 cm². Each semicircle has a diameter of 21 cm, so the radius is 10.5 cm. The area of each semicircle works out to (22/7) × (10.5)² / 2 = (22 × 10.5 × 10.5) / 14 = 2425.5 / 14 = 173.25 cm². The total shaded region includes both semicircles plus the two triangles: (2 × 173.25) + 220.50 = 346.50 + 220.50 = 567 cm².
In simple words: The shaded region covers both semicircles and two of the four triangles made by the diagonals. Add up all these pieces to get the final area.

Exam Tip: Remember that the diagonals of any square always create four equal triangles - this fact is key to solving many shaded region problems quickly.

Question 19(a). In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.
Answer: The square has an area of 14² = 196 cm². The diameter of each semicircle matches the side of the square, which is 14 cm, giving a radius of 7 cm. The combined area of both semicircles is 2 × (πr²/2) = 2 × [(22/7) × 49 / 2] = (22 × 49) / 7 = 22 × 7 = 154 cm². The shaded region is found by subtracting the semicircles from the square: 196 - 154 = 42 cm². For the perimeter, we add the arc of the first semicircle, the top side of the square, the arc of the second semicircle, and the bottom side: πr + 14 + πr + 14 = 2πr + 28 = 2 × (22/7) × 7 + 28 = 44 + 28 = 72 cm.
In simple words: Subtract the total area of both semicircles from the square to find the shaded area. The perimeter includes two curved arcs plus two straight sides.

Exam Tip: Always carefully identify which parts of the figure form the boundary of the shaded region - missing just one arc or side will cost marks in the perimeter calculation.

Question 19(b). In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.
Answer: Let the radius of each circle be r cm. From the figure, four radii fit along the side length: r + r + r + r = 14, which means 4r = 14, so r = 3.5 cm. Since all four circles have the same radius, they all have the same area. Each circle's area is πr² = (22/7) × (3.5)² = (22/7) × 3.5 × 3.5 = 22 × 0.5 × 3.5 = 38.5 cm². The total area covered by all four circles is 4 × 38.5 = 154 cm². The shaded region (the spaces between the circles) equals the square's area minus the circles' total area: 196 - 154 = 42 cm².
In simple words: The shaded parts are the corners and gaps left over after four equal circles are placed in the square. Find each circle's size, add them up, then subtract from the square.

Exam Tip: Notice that the four circles are arranged so their diameters add up to the square's side - use this relationship to find the radius quickly without extra calculation.

Question 19(c). In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = 22/7.
Answer: The semicircle has a diameter of 14 cm (labeled as BD), so the radius is 7 cm. Let AF = FE = x. From the figure, these two segments make up the full diameter: x + x = 14, giving x = 7 cm. The area of the semicircle BCD is (πr²)/2 = (22/7) × 49 / 2 = (22 × 7) / 2 = 77 cm². Each quadrant (ABF and EDF) has an area of (πr²)/4 = (22/7) × 49 / 4 = 154/4 = 38.5 cm². The rectangle ABDE has dimensions 7 × 14 = 98 cm². The shaded region is calculated as: rectangle area plus semicircle area minus both quadrants: 98 + 77 - 38.5 - 38.5 = 98 cm².
In simple words: Start with the rectangle, add the large curved bottom section, then subtract the two curved corners on top. These steps combine to give the total shaded area.

Exam Tip: Breaking a complex figure into simpler pieces (rectangle, semicircle, quadrants) makes the calculation manageable - always look for ways to split irregular shapes into standard ones.

Question 20(a). Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimeters.
Answer: The larger semicircle has a radius of 14 cm. Its area is (πR²)/2 = (22/7) × 196 / 2 = (22 × 196) / 14 = 22 × 14 = 308 cm². The smaller semicircle has a diameter of 14 cm, so its radius is 7 cm, and its area is (πr²)/2 = (22/7) × 49 / 2 = (22 × 49) / 14 = 11 × 7 = 77 cm². The shaded area is the difference: 308 - 77 = 231 cm². For the perimeter, we need the arc of the large semicircle, the arc (or circumference) of the small semicircle, and the straight segment connecting them: πR + πr + 14 = (22/7) × 14 + (22/7) × 7 + 14 = 44 + 22 + 14 = 80 cm.
In simple words: Imagine scooping out a smaller semicircle from a larger one. The shaded region is what remains. The perimeter traces along the outer curve, the inner curve, and the straight base.

Exam Tip: For concentric or nested shapes, always subtract the inner area from the outer area - and for perimeter, make sure you trace the actual boundary, not a line that goes through the interior.

Question 20(b). In the figure (ii) given below, area of △ABC = 35 cm². Find the area of the shaded region.
Answer: The area formula for a triangle is (1/2) × base × height. Given that △ABC has an area of 35 cm² and the height CD is 5 cm, we can find the base: (1/2) × AB × 5 = 35, which gives AB = (35 × 2) / 5 = 14 cm. The semicircle has AB as its diameter, so the radius is 7 cm. The semicircle's area works out to (πr²)/2 = (22/7) × 49 / 2 = (22 × 49) / 14 = 11 × 7 = 77 cm². The shaded region is the part of the semicircle not covered by the triangle: 77 - 35 = 42 cm².
In simple words: The semicircle sits on top of the triangle's base. The shaded part is the curved area left after the triangle is removed from the semicircle.

Exam Tip: When a triangle and a curved shape overlap, work backwards from the given triangle area to find the base or height - this often unlocks the radius or diameter you need.

Question 21(a). In the figure (i) given below, AOBC is a quadrant of a circle of radius 10 m. Calculate the area of the shaded portion. Take π = 3.14 and give your answer correct to two significant figures.
Answer: A quadrant is one-quarter of a circle. The area of quadrant AOBC is (πr²)/4 = (3.14 × 100) / 4 = 314 / 4 = 78.5 m². The right triangle AOB has two perpendicular sides (the radii) of 10 m each. Its area is (1/2) × 10 × 10 = 50 m². The curved region (shaded area) is found by subtracting the triangle from the quadrant: 78.5 - 50 = 28.5 m². Rounding to two significant figures gives 29 m² (or 28 m² if rounding down, depending on the convention; 28.5 rounds to 28 using banker's rounding, but typically 29 is expected here).
In simple words: A quadrant looks like a pizza slice with a quarter-circle curve. Remove the triangular part to see just the curved shaded region at the top.

Exam Tip: Always check the instruction for rounding - "two significant figures" means you keep two non-zero digits, so 28.5 becomes 29 (or 28 with banker's rounding). State your final answer clearly with units.

Question 21(b). In the figure (ii) given below, OAB is a quadrant of a circle. The radius OA = 7 cm and OD = 4 cm. Calculate the area of the shaded portion.
Answer: The area of a quadrant is found using the formula \( \frac{\pi r^2}{4} \). With radius 7 cm, the quadrant area becomes \( \frac{22}{7} \times \frac{7^2}{4} = \frac{22 \times 7}{4} = 38.5 \text{ cm}^2 \). For the triangle AOD, we use the formula \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 4 = 14 \text{ cm}^2 \). The shaded region is obtained by subtracting the triangle's area from the quadrant's area: \( 38.5 - 14 = 24.5 \text{ cm}^2 \).
In simple words: Find the quarter-circle area, then find the triangle area, and take away the triangle from the quarter-circle to get the shaded part.

Exam Tip: Always identify which shape is the quadrant and which is the triangle in the figure - this determines which area you subtract from which.

Question 22. A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle.
Answer: The rectangular paper measures 30 cm by 21 cm, giving a total area of 630 cm². The largest circle that fits inside this rectangle will have a diameter equal to the shorter side, which is 21 cm. Therefore, the radius is 10.5 cm. Using the area formula, the circle's area is \( \frac{22}{7} \times (10.5)^2 = \frac{22}{7} \times 110.25 = 346.5 \text{ cm}^2 \). The remaining paper area is obtained by subtracting the circle from the rectangle: \( 630 - 346.5 = 283.5 \text{ cm}^2 \).
In simple words: The biggest circle that fits has width equal to the shorter edge of the rectangle. Subtract the circle's area from the rectangle's area to find leftover paper.

Exam Tip: Remember that the diameter of the largest inscribed circle equals the shorter dimension of the rectangle - this is a common question pattern.

Question 23. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.
Answer: When a rectangle is inscribed in a circle, its diagonal equals the circle's diameter. The diameter is \( 2 \times 2.5 = 5 \) cm. One side of the rectangle is given as 4 cm. Using the Pythagorean theorem for the right triangle formed by two adjacent sides and the diagonal: \( AC^2 = AB^2 + BC^2 \).
\[ 5^2 = 4^2 + BC^2 \]
\[ 25 = 16 + BC^2 \]
\[ BC^2 = 9 \]
\[ BC = 3 \text{ cm} \]
The area of the rectangle is \( 4 \times 3 = 12 \text{ cm}^2 \).
In simple words: The diagonal of the rectangle is the circle's diameter. Once you know one side and the diagonal, use the Pythagorean theorem to find the other side, then multiply to get area.

Exam Tip: In problems involving shapes inscribed in circles, the diagonal (or longest dimension) always equals the diameter - use this relationship to set up your equation.

Question 24(a). In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3.142)
Answer: The rectangle inscribed in the circle has dimensions 12 cm by 5 cm. To find the circle's size, we calculate the diagonal (which is the diameter): \( AC^2 = 5^2 + 12^2 = 25 + 144 = 169 \), so \( AC = 13 \) cm. The radius is \( \frac{13}{2} = 6.5 \) cm. The circle's area is \( 3.142 \times (6.5)^2 = 3.142 \times 42.25 = 132.75 \text{ cm}^2 \). The rectangle's area is \( 12 \times 5 = 60 \text{ cm}^2 \). The shaded region (between circle and rectangle) is \( 132.75 - 60 = 72.75 \text{ cm}^2 \).
In simple words: Find the circle's radius using the rectangle's diagonal. Calculate both areas, then subtract the rectangle from the circle.

Exam Tip: Always round your final answer to the number of decimal places requested - here, two decimal places means you stop at the hundredths position.

Question 24(b). In the figure (ii) given below, ABC is an isosceles right angled triangle with \( \angle ABC = 90° \). A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take \( \pi = \frac{22}{7} \).
Answer: The triangle ABC is isosceles and right-angled, with both legs measuring 7 cm. The area of the triangle is \( \frac{1}{2} \times 7 \times 7 = 24.5 \text{ cm}^2 \). Using the Pythagorean theorem, the hypotenuse AC is \( \sqrt{7^2 + 7^2} = \sqrt{98} = 7\sqrt{2} \) cm. This is the diameter of the semicircle, so the radius is \( \frac{7\sqrt{2}}{2} \) cm. The semicircle's area is \( \frac{1}{2} \times \frac{22}{7} \times \left(\frac{7\sqrt{2}}{2}\right)^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{98}{4} = \frac{2156}{56} = 38.5 \text{ cm}^2 \). The shaded region (between semicircle and triangle) is \( 38.5 - 24.5 = 14 \text{ cm}^2 \).
In simple words: Find the triangle area using both legs. Find the hypotenuse using Pythagorean theorem - this is the semicircle's diameter. Calculate the semicircle's area, then subtract the triangle.

Exam Tip: When working with isosceles right triangles, remember that the hypotenuse equals \( \text{side} \times \sqrt{2} \) - this shortcut saves time and reduces errors.

Question 25. A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.
Answer: From the perimeter, we find the radius: \( 2\pi r = 660 \), so \( \pi r = 330 \). With \( \pi = \frac{22}{7} \), we get \( r = \frac{330 \times 7}{22} = \frac{2310}{22} = 105 \) m. When a square is inscribed in a circle with vertices on the circumference, the diagonal of the square equals the diameter of the circle. The diameter is \( 2 \times 105 = 210 \) m. For a square with side \( x \), the diagonal is \( x\sqrt{2} \). Setting \( x\sqrt{2} = 210 \), we get \( x^2 = \frac{210^2}{2} = \frac{44100}{2} = 22050 \text{ m}^2 \).
In simple words: Find the radius from the circle's perimeter. The diagonal of the inscribed square equals the circle's diameter. Use this to find the square's side, then calculate its area.

Exam Tip: For a square inscribed in a circle, the relationship diagonal = diameter is key - squaring the side gives area directly without needing to compute the side length separately.

Question 26. In the adjoining figure, ABCD is a square. Find the ratio between
(i) the circumferences
(ii) the areas of the incircle and the circumcircle of the square.
Answer: Let the side of the square be \( 2a \) units.

(i) Circumferences: The incircle touches all four sides of the square, so its diameter equals the side length of the square. The radius of the incircle is \( r = a \) units. The circumcircle passes through all four vertices of the square, so its diameter is the diagonal of the square. The diagonal is \( AC = \sqrt{(2a)^2 + (2a)^2} = \sqrt{8a^2} = 2\sqrt{2}a \) units, giving a circumradius of \( R = \sqrt{2}a \) units. The circumference of the incircle is \( 2\pi r = 2\pi a \). The circumference of the circumcircle is \( 2\pi R = 2\pi\sqrt{2}a \). The ratio is \( \frac{2\pi a}{2\pi\sqrt{2}a} = \frac{1}{\sqrt{2}} = 1 : \sqrt{2} \).

(ii) Areas: The area of the incircle is \( \pi r^2 = \pi a^2 \). The area of the circumcircle is \( \pi R^2 = \pi(\sqrt{2}a)^2 = 2\pi a^2 \). The ratio is \( \frac{\pi a^2}{2\pi a^2} = \frac{1}{2} = 1 : 2 \).
In simple words: The incircle sits inside the square touching all sides, while the circumcircle goes through all four corners. Their radii differ by a \( \sqrt{2} \) factor because the diagonal is \( \sqrt{2} \) times the side length.

Exam Tip: Always set up the square's side as an algebraic term (like 2a) to make calculations cleaner - the final ratio should simplify nicely and not depend on the actual side length chosen.

Question 27(a). The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end. PQ = 200 m; PT = 70 m.
(i) Calculate the area of the grassed enclosure in m².
(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.
Answer: (i) Using the formula for the area of a rectangle, we get area of PQST = 200 × 70 = 14000 m². The radius of each semicircular part on either side = PT/2 = 70/2 = 35 m. Area of both semicircular parts = 2 × (πr²/2) = πr² = (22/7) × 35 × 35 = 22 × 5 × 35 = 3850 m². So the total area of the grassed enclosure = 14000 + 3850 = 17850 m².

(ii) The width of the track around the enclosure = 7 m. From the figure, AB = PQ = 200 m and ED = ST = 200 m. Also, EA = PT + ET + AP = 70 + 7 + 7 = 84 m, and BD = DS + QS + BQ = 70 + 7 + 7 = 84 m. The outer radius of the semicircle (R) = EA/2 = 84/2 = 42 m. Circumference of both semicircular parts = 2πR = 2 × (22/7) × 42 = 264 m. Therefore, the outer perimeter = circumference of both semicircular parts + ED + AB = 264 + 200 + 200 = 664 m.
In simple words: The grassed area equals the rectangle plus two semicircles. For the outer track, add the outer curved parts and two straight sides.

Exam Tip: Always identify the radius for each semicircular region carefully - for the inner enclosure, it's half the width of the rectangle, and for the outer track, add the track width to get the outer radius.

Question 27(b). In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.
Answer: Given that the inner perimeter = 312 m and straight portion = 90 m. Setting up the equation: 90 + πr + 90 + πr = 312, which simplifies to 2πr + 180 = 312. Solving: 2πr = 132, so πr = 66. This gives r = 66 ÷ (22/7) = (66 × 7)/22 = 21 m. Therefore, the length of AB = 2r = 2 × 21 = 42 m. Since the track width = 2 m, the outer length HE = GF = 42 + 2 + 2 = 46 m. The radius of the outer semicircle (R) = 46/2 = 23 m. The area of the track is calculated as: Area of outer semicircles + area of outer rectangle - (area of inner semicircles + area of inner rectangle) = πR² + 90 × 46 - (πr² + 42 × 90) = π(R² - r²) + 90(46 - 42) = (22/7)(23² - 21²) + 90 × 4 = (22/7)(529 - 441) + 360 = (22/7) × 88 + 360 = 1936/7 + 360 = (1936 + 2520)/7 = 4456/7 = 636 4/7 m².
In simple words: Find the inner radius from the perimeter equation. Then add the track width to get the outer radius. The shaded track area equals the difference between the outer and inner shapes.

Exam Tip: When working with concentric shapes (one inside the other), always calculate the outer and inner dimensions separately, then subtract to find the width or area between them.

Question 28(a). In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.
Answer: The radius of the circle with centre A equals AC = 8 cm. Using the formula, area of circle with centre A = πr² = (22/7) × 8² = (22/7) × 64 = 1408/7 = 201.14 cm². From the figure, BC = AC - AB = 8 - 3 = 5 cm. The area of the circle with centre B = πr² = (22/7) × 5² = (22/7) × 25 = 550/7 = 78.57 cm². The area of the shaded region = area of larger circle - area of smaller circle = 201.14 - 78.57 = 122.57 cm².
In simple words: The shaded part is the region where the larger circle extends beyond the smaller circle. Subtract the smaller area from the larger area to find the shaded portion.

Exam Tip: When two circles touch each other, the radius of the smaller circle can be found by subtracting the distance between centres from the radius of the larger circle.

Question 28(b). The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.
Answer: Each quadrant has a radius of 7 cm. From the figure, the shaded region equals the area of the square minus the area of all four quadrants. The side of the square = 2 × 7 = 14 cm. Area of square = 14² = 196 cm². Area of 4 quadrants = 4 × (1/4)πr² = πr² = (22/7) × 7 × 7 = 22 × 7 = 154 cm². Therefore, the area of the shaded region = 196 - 154 = 42 cm².
In simple words: A square contains four curved sections at its corners. The shaded part in the middle is what remains when you take away these curved pieces from the total square.

Exam Tip: When quadrants are placed at the corners of a square, four quarter-circles together equal one full circle - use this to simplify calculations.

Question 29(a). In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.
Answer: The area of square lawn ABCD = (side)² = 56² = 3136 m². In a square, the length of the diagonal = side√2 = 56√2 m. Since the diagonals bisect each other and are equal, we have AO = OC = OD = OB. The radius of each quadrant ODC and OAB = AC/2 = (56√2)/2 = 28√2 m. Area of each quadrant = (1/4)πr² = (1/4) × (22/7) × (28√2)² = (1/4) × (22/7) × 28√2 × 28√2 = 22 × 28 × 2 = 1232 m². The square is divided into four equal triangles by its diagonals. Area of each triangle = (area of square)/4 = 3136/4 = 784 m². The area of each flower bed = area of quadrant - area of triangle = 1232 - 784 = 448 m². Since there are two flower beds, total flower bed area = 2 × 448 = 896 m². Therefore, the sum of areas of lawn and flower beds = 3136 + 896 = 4032 m².
In simple words: The flower beds are curved sections centred at the middle of the square. Each one is a quadrant minus a triangle. Add all these areas plus the square itself.

Exam Tip: When diagonals divide a square, each resulting triangle has area equal to 1/4 of the square. Use this property to find areas quickly.

Question 29(b). In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (π = 3.14)
Answer: Given that each side of square OABC = 20 cm. The diagonal of a square = side√2 = 20√2 cm. Therefore, OB = 20√2 cm. Since OB is the radius of quadrant OPBQ, the area of the quadrant = (1/4)πr² = (1/4) × 3.14 × (20√2)² = (1/4) × 3.14 × 20√2 × 20√2 = 3.14 × 200 = 628 cm². The area of square OABC = 20² = 400 cm². The area of the shaded region = area of quadrant - area of square = 628 - 400 = 228 cm².
In simple words: The square sits inside a curved quadrant section. The shaded area is the curved part that sticks out beyond the square's corner.

Exam Tip: In problems with inscribed shapes, always find the radius or boundary of the outer shape using the diagonal of the inner square or rectangle.

Question 30(a). In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion.
Answer: Since ABCD is a rectangle, AD = BC = 7 cm and CD = AB = 14 cm. From the figure, a semicircle is drawn on DC (the top side) with diameter 14 cm, and two semicircles are drawn on BC and AD (the side sides) each with diameter 7 cm. The radius of the semicircle on DC = 14/2 = 7 cm, and the radius of each semicircle on BC and AD = 7/2 = 3.5 cm. Area of semicircle on DC = (1/2)πr² = (1/2) × (22/7) × 7² = (1/2) × 22 × 7 = 77 cm². Area of each semicircle on BC and AD = (1/2)πr² = (1/2) × (22/7) × (3.5)² = (1/2) × (22/7) × 12.25 = 19.25 cm². Area of rectangle ABCD = 14 × 7 = 98 cm². The shaded portion consists of the semicircle on DC and the rectangle, minus the two semicircles on BC and AD. Shaded area = 77 + 98 - (19.25 + 19.25) = 77 + 98 - 38.5 = 136.5 cm².
In simple words: Three curved pieces are drawn on the rectangle's edges. The shaded area equals the rectangle plus the top curve, minus the two side curves.

Exam Tip: Always identify which semicircles are to be added (inside the shaded region) and which are to be subtracted (outside the shaded region) by carefully studying the figure.

Question 30(b). In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14)
Answer: Since an angle in a semi-circle measures 90°, we know ∠A = 90°. In the right-angled triangle ABC, applying the Pythagoras theorem gives us BC² = AC² + AB² = 24² + 7² = 576 + 49 = 625, so BC = 25 cm. The circle's radius equals half the diameter BC, which is 12.5 cm. The triangle ABC has an area of (1/2) × 7 × 24 = 84 cm². The total area of the circle is π × 12.5² = 3.14 × 156.25 = 490.63 cm². The quadrant COD covers (1/4) × 490.63 = 122.66 cm². Therefore, the shaded area is 490.63 - (84 + 122.66) = 490.63 - 206.66 = 283.97 cm².
In simple words: The angle at A is 90 degrees. Use the Pythagoras rule to find BC = 25 cm. The radius is 12.5 cm. Subtract the triangle and quadrant areas from the circle's total area.

Exam Tip: Always verify that the angle inscribed in a semi-circle is 90° - this is a key property that unlocks the geometry. Check your quadrant calculation carefully, as it directly affects the final answer.

Question 31(a). In the figure (i) given below, ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circles which touch externally in pairs. Find the area of the shaded region.
Answer: Let r represent the radius of each circle. Since the circles touch externally and their centres are at the corners of the square, the distance between adjacent centres (one side of the square) equals 2r. Thus 2r = 14, giving r = 7 cm. Each complete circle has area π × 7² = (22/7) × 49 = 154 cm². The total area of all 4 circles is 4 × 154 = 616 cm². Each quadrant (quarter-circle) inside the square has area (1/4) × π × 7² = 38.5 cm², and 4 quadrants together cover 154 cm². The square's area is 14² = 196 cm². The shaded region consists of the parts of the circles outside the square, plus the part of the square not covered by quadrants. This equals (616 - 154) + (196 - 154) = 462 + 42 = 504 cm².
In simple words: The four circles touch at the corners. Their radius is 7 cm. Add the circle area outside the square to the square area outside the circles to get the shaded total.

Exam Tip: Recognize that the four corner quadrants together form one complete circle inside the square. Subtracting this once from both the 4 circles and the square avoids double-counting errors.

Question 31(b). In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semicircular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate. (i) the length of the boundary. (ii) the area of the shaded region. (Take π to be 3.14)
Answer:
(i) The big semi-circle has diameter 10 cm and radius R = 5 cm. The diameter of each smaller semi-circle is 5 cm, so each has radius r = 2.5 cm. The perimeter is the arc of the big semi-circle plus the arcs of both small semi-circles: πR + πr + πr = π(R + 2r) = 3.14(5 + 5) = 3.14 × 10 = 31.4 cm.
(ii) The shaded area equals the big semi-circle plus one small semi-circle minus the other small semi-circle, which simplifies to just the big semi-circle's area: (1/2) × π × 5² = (1/2) × 3.14 × 25 = 1.57 × 25 = 39.25 cm².
In simple words: The boundary goes around three arcs. The smaller ones have the same size. For area, the two small semi-circles cancel out, leaving only the big one.

Exam Tip: Pay careful attention to which arcs form the boundary - only count the outer curves. When calculating shaded area, notice that adding and subtracting equal regions simplifies the formula dramatically.

Question 32(a). In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cm and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take π = 3.14)
Answer: Let r₁ = 5 cm, r₂ = 3 cm, and r₃ = 2 cm. The perimeter of the shaded region is the sum of the three semi-circle arcs: πr₁ + πr₂ + πr₃ = π(5 + 3 + 2) = 10π = 10 × 3.14 = 31.4 cm. For the area, the shaded portion equals the largest semi-circle minus the middle one, plus the smallest one: (πr₁²/2) - (πr₂²/2) + (πr₃²/2) = (π/2)(r₁² - r₂² + r₃²) = (3.14/2)(25 - 9 + 4) = (3.14/2) × 20 = 3.14 × 10 = 31.4 cm².
In simple words: Three arcs make the edge, so add all their lengths. The shaded area alternates between adding and subtracting semi-circles.

Exam Tip: When regions are nested concentrically, use alternating addition and subtraction to find the shaded part. Always double-check your sign pattern matches the figure.

Question 32(b). In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the area of the shaded region. Take π = 3.14.
Answer: The square has side 4 cm and area 16 cm². Each of the four corner quadrants has radius r = 1 cm and area (1/4)πr² = (1/4) × 3.14 × 1 = 0.785 cm² (total for 4: 3.14 cm²). The central circle has diameter 2 cm, so radius r₁ = 1 cm and area πr₁² = 3.14 × 1 = 3.14 cm². The shaded area is the part of the square not covered by these regions: 16 - 3.14 - 3.14 = 9.72 cm².
In simple words: Start with the full square. Remove the four corner quarter-circles and the middle circle. What's left is the shaded region.

Exam Tip: Verify that four quarter-circles of radius 1 cm exactly fill one complete circle of radius 1 cm - this insight can speed up your calculation. Always check that your subtractions don't overlap unintentionally.

Question 33(a). In the figure (i) given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle.
Answer: The rectangle's area is AB × BC = 14 × 7 = 98 cm². Since ABCD is a rectangle, CD = AB = 14 cm. The quarter-circle BFEC has radius BC = 7 cm, so CE = 7 cm. This means DE = CD - CE = 14 - 7 = 7 cm. The semi-circle DGE has diameter DE = 7 cm and radius 3.5 cm, giving area (1/2) × π × (3.5)² = (1/2) × (22/7) × 12.25 = 19.25 cm². The quarter-circle BFEC has radius 7 cm and area (1/4) × π × 7² = (1/4) × (22/7) × 49 = 38.5 cm². The remaining area is 98 - 19.25 - 38.5 = 40.25 cm².
In simple words: Start with the rectangle area. Subtract both the semi-circle and the quarter-circle that are removed from it.

Exam Tip: Identify the radius of each curved region carefully - the quadrant uses one side of the rectangle, while the semi-circle is based on the remainder. Label all intermediate lengths (CE, DE) to avoid mixing up which curve uses which dimension.

Question 33(b). The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and △CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.
Answer: Since ABCD is a square and the quadrant BCD has radius 42 cm, the side of the square is 42 cm, giving an area of 42² = 1764 cm². The quadrant BCD (a quarter of a circle with radius 42) has area (1/4) × π × 42² = (1/4) × (22/7) × 1764. Computing: (1/4) × (22/7) × 1764 = (22 × 441)/7 = 9702/7 = 1385.71 cm² (approximately 1386 cm²). The isosceles right triangle CEF has equal sides of 6 cm each, so its area is (1/2) × 6 × 6 = 18 cm². The shaded region depends on which parts are shaded in the figure. Typically, for a kite configuration, the shaded area would be the quadrant plus the triangle, or the square minus these regions - verify from the diagram. Assuming the shaded region is the quadrant plus the triangle: 1386 + 18 = 1404 cm² (or the remaining part of the square after both the quadrant and triangle are removed from consideration).
In simple words: Find the area of the square. Calculate the quadrant's area using the 42 cm radius. Find the triangle area from its two equal 6 cm sides. Combine these based on what the shading shows.

Exam Tip: For kite problems, always clarify which regions are shaded versus unshaded by referring carefully to the figure. The quadrant's large radius (42 cm) dominates the calculation - double-check your multiplication and division steps for accuracy.

Question 34(a). In the figure (i) given below, the boundary of the shaded region in the given diagram consists of four semicircular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, calculate (i) the length of the boundary. (ii) the area of the shaded region.
Answer: (i) From the figure, we find that OA = 14/2 = 7 cm. Since OB = OA - AB = 7 - 3.5 = 3.5 cm, the radius of the smallest semi-circle is 3.5/2 = 1.75 cm. The circumference of a semi-circle equals πr. Therefore, the boundary length = Circumference of largest semi-circle + Circumference of smaller semi-circle + 2 × Circumference of smallest semi-circle = 7π + 3.5π + (2 × 1.75π) = 7π + 3.5π + 3.5π = 14π = 14 × 22/7 = 2 × 22 = 44 cm.

(ii) The shaded area = Area of large semi-circle + Area of smaller semi-circle - 2 × Area of smallest semi-circle. This gives us [π(7)²/2] + [π(3.5)²/2] - [2 × π(1.75)²/2] = [49π/2] + [12.25π/2] - [6.125π/2] = [55.125π/2] = [55.125 × 22/7]/2 = 1212.75/14 = 86.625 cm².
In simple words: Add up the lengths of all four curved parts that form the boundary. For the shaded area, calculate the big semicircle's area, add the medium semicircle's area, then subtract twice the small semicircles' areas.

Exam Tip: Always identify which arcs form the boundary (those you add) versus which you subtract. Double-check that you're using radius correctly - the two smallest semicircles appear twice in the perimeter calculation.

Question 34(b). In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD and AB || CD and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.
Answer: From the figure, the radius of the quadrant is BC = 3.5 cm. Since both EC and the radius equal 3.5 cm (EC is also a radius of quadrant BFEC), we use the trapezium formula: Area = (1/2) × (Sum of parallel sides) × Distance between them = (1/2) × (AB + EC + DE) × BC = (1/2) × (3.5 + 3.5 + 2) × 3.5 = (1/2) × 9 × 3.5 = 4.5 × 3.5 = 15.75 cm². The quadrant's area equals πr²/4 = (22/7) × (3.5)²/4 = (22/7) × 12.25/4 = (22 × 12.25)/(7 × 4) = 269.5/28 = 9.625 cm². Therefore, the shaded portion = Area of trapezium - Area of quadrant = 15.75 - 9.625 = 6.125 cm².
In simple words: First find the trapezium's area using the lengths of its parallel sides. Then subtract the quarter circle's area to get what remains.

Exam Tip: Remember that when a quarter circle is removed, you must first identify all the trapezium's dimensions correctly - use the given measurements along with the radius relationship to find missing lengths.

Question 35(a). In the figure (i) given below, ABC is a right angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
Answer: In the right angle triangle ABC, by Pythagoras' theorem: AC² = AB² + BC² = 28² + 21² = 784 + 441 = 1225, so AC = 35 cm. The radius of the semi-circle (R) = AC/2 = 35/2 = 17.5 cm, and the radius of the quadrant (r) = BC = 21 cm. From the figure, the shaded area = Area of triangle ABC + Area of semi-circle - Area of quadrant. This equals (1/2) × BC × AB + πR²/2 - πr²/4 = (1/2) × 21 × 28 + (22/7) × (17.5)²/2 - (22/7) × (21)²/4 = 21 × 14 + (22 × 306.25)/(7 × 2) - (22 × 441)/(7 × 4) = 294 + 6737.5/14 - 9702/28 = 294 + 481.25 - 346.5 = 428.75 cm².
In simple words: Use the Pythagorean theorem to find the hypotenuse AC. Then add the triangle and semicircle areas, then subtract the quarter circle area.

Exam Tip: Ensure you use Pythagoras correctly and don't mix up which measurement (AB or BC) serves as the base versus the height for the triangle area calculation.

Question 35(b). In the figure (ii) given below, ABC is an equilateral triangle of side 8 cm. A, B and C are the centers of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places.
Answer: Since triangle ABC is equilateral with side 8 cm, and A, B, C are centres of three circular arcs of equal radius, the radius of each arc is 8/2 = 4 cm. Using the formula for the area of an equilateral triangle: Area = (√3/4) × side² = (√3/4) × 8 × 8 = (√3/4) × 64 = 16√3 = 16 × 1.732 = 27.712 cm². The area of three equal sectors of 60° (since each central angle is 360°/6 = 60°) whose radius is 4 cm equals 3 × πr² × (60°/360°) = 3 × 3.142 × 4 × 4 × (1/6) = 3.142 × 8 = 25.136 cm². The shaded region = Area of equilateral triangle - Area of 3 sectors = 27.712 - 25.136 = 2.576 ≈ 2.58 cm².
In simple words: Find the triangle's area first. Then work out the three sector areas and subtract them to see what remains as shaded.

Exam Tip: The central angle at each vertex of an equilateral triangle is 60 degrees - use this relationship when calculating sector areas. Round your final answer to two decimal places as required.

Question 36. A circle is inscribed in a regular hexagon of side 2√3 cm. Find (i) the circumference of the inscribed circle (ii) the area of the inscribed circle
Answer: (i) The side of the regular hexagon (a) = 2√3 cm. Since a regular hexagon has 6 sides, the central angle = 360°/6 = 60°. In triangle AOB, since ∠AOB = 60° and the sum of all angles in a triangle is 180°, we have ∠AOB + ∠OAB + ∠OBA = 180°. Since OA = OB (both are radii), the base angles are equal: ∠OAB = ∠OBA = x. Substituting, 60° + x + x = 180°, which gives 2x = 120°, so x = 60°. Since all three angles equal 60°, triangle AOB is equilateral, meaning AO = BO = AB = 2√3 cm. Drawing a perpendicular from O to AB, we get AT = BT = (2√3)/2 = √3 cm (since in an equilateral triangle, altitude and median coincide). In right angle triangle OAT: OA² = OT² + AT², so (2√3)² = OT² + (√3)², giving 12 = OT² + 3, thus OT² = 9, and OT = 3 cm. The radius of the inscribed circle = 3 cm. The circumference = 2πr = 2 × (22/7) × 3 = 132/7 cm.

(ii) Area of inscribed circle = πr² = (22/7) × 3² = (22/7) × 9 = 198/7 cm².
In simple words: A hexagon's central angle splits into 60 degrees. The triangle formed is equilateral, so you can use its properties to find how far the circle's centre is from each side. That distance is the circle's radius.

Exam Tip: Recognizing that triangle AOB is equilateral unlocks the problem - don't just treat it as a generic isosceles triangle. Use the perpendicular to the base to apply the Pythagorean theorem cleanly.

Question 37. In the adjoining figure, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.
Answer: Given: Radius of the circle = 10 cm, and the angle at the centre subtended by chord AB = 90°. The area of sector OACB = πr² × (θ/360°) = 3.14 × 10 × 10 × (90°/360°) = 3.14 × 100 × (1/4) = 314/4 = 78.5 cm². For the major segment, we first find the area of triangle OAB. Since ∠AOB = 90° and OA = OB = 10 cm (radii), triangle OAB is a right isosceles triangle. Its area = (1/2) × OA × OB = (1/2) × 10 × 10 = 50 cm². The minor segment area = Sector area - Triangle area = 78.5 - 50 = 28.5 cm². The total area of the circle = πr² = 3.14 × 100 = 314 cm². The major segment area = Total circle area - Minor segment area = 314 - 28.5 = 285.5 cm².
In simple words: The sector is a slice of the circle from the centre. To find the major segment, subtract the minor segment (which includes the triangle) from the whole circle.

Exam Tip: Always identify whether you need the minor or major segment - the minor segment is smaller and lies on one side of the chord, while the major segment is everything else in the circle. Use the sector formula with the correct angle in degrees.

Exercise 15.4

Question 1. Find the surface area and volume of a cube whose one edge is 7 cm.
Answer: The edge of the cube is 7 cm. Using the surface area formula, we get 6a² = 6 × (7)² = 6 × 7 × 7 = 294 cm². For the volume, we use the formula a³ = (7)³ = 7 × 7 × 7 = 343 cm³. So the surface area is 294 cm² and the volume is 343 cm³.
In simple words: A cube with sides of 7 cm has a surface area of 294 cm² and takes up a space of 343 cm³.

Exam Tip: Remember that surface area uses a² (six faces) while volume uses a³ (the space inside). Always check your units - area is squared, volume is cubed.

Question 2. Find the surface area and the volume of a rectangular solid measuring 5 m by 4 m by 3 m. Also find the length of a diagonal.
Answer: For the rectangular solid with length 5 m, breadth 4 m, and height 3 m, we apply the surface area formula 2(lb + bh + lh) = 2(5 × 4 + 4 × 3 + 5 × 3) = 2(20 + 12 + 15) = 2 × 47 = 94 m². The volume is l × b × h = 5 × 4 × 3 = 60 m³. For the diagonal, we use the formula \( \sqrt{l^2 + b^2 + h^2} = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2} = 5 × 1.414 = 7.07 \) m.
In simple words: The box covers 94 m² on the outside, holds 60 m³ inside, and the longest stick you can fit through it measures 7.07 m.

Exam Tip: The diagonal of a box is the longest distance from one corner to the opposite corner. Always simplify square roots and use approximate values only at the final step.

Question 3. The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000 cm³, find its height.
Answer: Given that the length is 25 cm, the breadth is 20 cm, and the volume is 7000 cm³. We need to find the height. Using the volume formula l × b × h = 7000, we substitute the known values: 25 × 20 × h = 7000, which gives 500 × h = 7000. Solving for h: h = 7000 ÷ 500 = 14 cm.
In simple words: When you know the volume and two measurements of a box, divide the volume by the product of those two measurements to find the missing measurement.

Exam Tip: Always rearrange the formula to isolate the unknown variable. Check your answer by multiplying back: 25 × 20 × 14 should equal 7000.

Question 4. A class room is 10 m long, 6 m broad and 4 m high. How many students can it accommodate if one student needs 1.5 m² of floor area? How many cubic metres of air will each student have?
Answer: The classroom has length 10 m, breadth 6 m, and height 4 m. The floor area is l × b = 10 × 6 = 60 m². Since each student needs 1.5 m² of floor space, the number of students is 60 ÷ 1.5 = 40 students. The volume of the classroom is l × b × h = 10 × 6 × 4 = 240 m³. The air available per student is 240 ÷ 40 = 6 m³.
In simple words: Forty students can fit in the room, and each student gets 6 cubic metres of air to breathe.

Exam Tip: Floor area is a 2D measurement (length × breadth), while air space is a 3D measurement (volume). Always match the unit to the dimension being measured.

Question 5(a). The volume of a cuboid is 1440 cm³. Its height is 10 cm and the cross-section is a square. Find the side of the square.
Answer: Given that the volume is 1440 cm³, the height is 10 cm, and the cross-section is a square, let the side of the square be x cm. The volume formula for a cuboid is volume = area of cross-section × height. So, 1440 = x² × 10. This gives x² = 1440 ÷ 10 = 144. Taking the square root, x = \( \sqrt{144} \) = 12 cm.
In simple words: When the cross-section is a square and you know the volume and height, find the side length by working backwards from the volume formula.

Exam Tip: Always check that your answer makes sense by substituting it back into the original formula: 12 × 12 × 10 = 1440 ✓

Question 5(b). The perimeter of one face of a cube is 20 cm. Find the surface area and the volume of the cube.
Answer: The perimeter of one face of a cube is 4 × side = 20 cm, so the side length is 20 ÷ 4 = 5 cm. The surface area is 6(side)² = 6 × (5)² = 6 × 5 × 5 = 150 cm². The volume is side³ = 5 × 5 × 5 = 125 cm³.
In simple words: The perimeter tells us the side length, which then helps us find both the outer area and the inner space.

Exam Tip: A cube has all equal sides. If you know the perimeter of one square face, you can find the side by dividing by 4 (since a square has 4 equal sides).

Question 6. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured papers with pictures of Santa Claus. She must know the exact quantity of paper to buy for this purpose. If the box has length 80 cm, breadth 40 cm and height 20 cm respectively, then how many square sheets of paper of side 40 cm would she require?
Answer: The surface area of the box is 2(lh + bh + lh) = 2(80 × 40 + 40 × 20 + 20 × 80) = 2(3200 + 800 + 1600) = 2 × 5600 = 11,200 cm². Each square sheet has area (40)² = 1600 cm². The number of sheets needed is 11,200 ÷ 1600 = 7 sheets.
In simple words: Calculate how much paper the box needs, divide by the area of each sheet, and you know how many sheets to buy.

Exam Tip: Make sure the units match. Both the box surface area and the sheet area should be in the same units (cm²) before dividing.

Question 7. The volume of a cuboid is 3600 cm³ and its height is 12 cm. The cross-section is a rectangle whose length and breadth are in the ratio 4 : 3. Find the perimeter of the cross-section.
Answer: Let the length be 4x cm and the breadth be 3x cm. Using the volume formula, 3600 = 4x × 3x × 12, which simplifies to 144x² = 3600. Solving for x²: x² = 3600 ÷ 144 = 25, so x = 5 cm. The length is 4 × 5 = 20 cm and the breadth is 3 × 5 = 15 cm. The perimeter is 2(l + b) = 2(20 + 15) = 2 × 35 = 70 cm.
In simple words: When sides are in a ratio, use variables with that ratio, substitute into the volume formula, solve for the variable, then find the actual dimensions.

Exam Tip: Always express dimensions in terms of a single variable when a ratio is given. This reduces the complexity of the equation and avoids mistakes.

Question 8. The volume of a cube is 729 cm³. Find its surface area and the length of a diagonal.
Answer: If the volume is 729 cm³, then (side)³ = 729, which means side = \( \sqrt[3]{729} \) = 9 cm. The surface area is 6(side)² = 6 × (9)² = 6 × 9 × 9 = 486 cm². The diagonal of a cube is \( \sqrt{3} \) × side = \( \sqrt{3} \) × 9 = 1.732 × 9 = 15.57 cm.
In simple words: Take the cube root of the volume to find the side length, then use that to calculate both the surface area and the diagonal.

Exam Tip: The diagonal of a cube passes through its interior from one corner to the opposite corner. The formula \( \sqrt{3} \) × side is unique to cubes and must be memorized.

Question 9. The length of the longest rod which can be kept inside a rectangular box is 17 cm. If the inner length and breadth of the box are 12 cm and 8 cm respectively, find its inner height.
Answer: The longest rod that fits inside a rectangular box equals the diagonal of the box. Using the diagonal formula \( \sqrt{l^2 + b^2 + h^2} \) = 17, we have \( \sqrt{12^2 + 8^2 + h^2} \) = 17. Squaring both sides: 144 + 64 + h² = 289, so 208 + h² = 289, which gives h² = 81. Taking the square root: h = 9 cm.
In simple words: The longest thing that fits inside a box goes from one corner to the opposite corner. Use this diagonal property and the given length to find the missing height.

Exam Tip: Always square both sides when dealing with square roots to eliminate the radical. Then rearrange to isolate the unknown squared term before taking the final square root.

Question 10. A closed rectangular box has inner dimensions 90 cm by 80 cm by 70 cm. Calculate its capacity and the area of tin-foil needed to line its inner surface.
Answer: The capacity (volume) is l × b × h = 90 × 80 × 70 = 504,000 cm³. The area of tin-foil needed is the surface area: 2(lb + bh + lh) = 2(90 × 80 + 80 × 70 + 90 × 70) = 2(7200 + 5600 + 6300) = 2 × 19,100 = 38,200 cm².
In simple words: Capacity is how much the box can hold (volume), while tin-foil area is how much you need to cover the inside (surface area).

Exam Tip: "Capacity" always refers to volume. When a problem mentions lining a surface, calculate the surface area. Read the problem carefully to identify which measurement is being asked for.

Question 11. The internal measurements of a box are 20 cm long, 16 cm wide and 24 cm high. How many 4 cm cubes could be put into the box?
Answer: The volume of the box is 20 × 16 × 24 = 7680 cm³. Each small cube has volume 4 × 4 × 4 = 64 cm³. The number of cubes that fit is 7680 ÷ 64 = 120 cubes.
In simple words: Divide the total volume of the box by the volume of one small cube to find how many small cubes fit inside.

Exam Tip: This is a straightforward division of volumes. Ensure both volumes are in the same units before dividing. Check your answer by multiplying: 120 × 64 should equal 7680.

Question 12. The internal measurements of a box are 10 cm long, 8 cm wide and 7 cm high. How many cubes of side 2 cm can be put into the box?
Answer: Since the box height is 7 cm, only 3 cubes fit when placed height-wise. (Placing 4 cubes would result in a height of 8 cm, exceeding the box height.) The total height we can use is 3 × 2 = 6 cm. Using the volume formula for a cuboid: Volume of box = 10 × 8 × 6 = 480 cm³. For each cube with side 2 cm: Volume of one cube = 2³ = 8 cm³. If n is the number of cubes, then 8n = 480, so n = 60. Therefore, 60 cubes can fit into the box.
In simple words: The box can only hold 3 cubes stacked up vertically. When you work out how much space they need and how much space each cube takes, you find that 60 cubes in total can fit inside.

Exam Tip: Always check the height constraint first - the limiting dimension often determines how many objects fit. Never assume all dimensions allow full coverage without verification.

Question 13. A certain quantity of wood costs Rs. 25,000 per m³. A solid cubical block of such wood is bought for Rs. 18,225. Calculate the volume of the block and use the method of factor to find the length of one edge of the block.
Answer: Given: Cost per m³ = Rs. 25,000 and total cost = Rs. 18,225. The volume can be found from: Volume = Total cost ÷ Cost per m³ = 18,225 ÷ 25,000 = 729/1000 = 0.729 m³. For a cube, Volume = (side)³, so (side)³ = 729/1000. Taking the cube root of both sides and factorizing: 729 = 3⁶ and 1000 = 2³ × 5³. Therefore, side = ∛(3⁶ / 2³ × 5³) = (3² × 3²) / (2 × 5) = 9/10 = 0.9 m. Hence, the volume is 0.729 m³ and the edge length is 0.9 m.
In simple words: Divide the total cost by the price per cubic metre to get the volume. Then find what number, when cubed (multiplied by itself three times), equals that volume - that gives you one edge length.

Exam Tip: Always show the complete factorization of both numerator and denominator before simplifying the cube root - examiners check that you understand how factors group into threes.

Question 14. A cube of 11 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base of the vessel are 15 cm × 12 cm, find the rise in the water level in centimeters correct to 2 decimal places, assuming that no water overflows.
Answer: Let the rise in water height be h cm. The volume of water displaced equals the volume of the cube. The base area of the vessel is 15 × 12 = 180 cm². The cube volume is 11³ = 1331 cm³. Setting up the equation: 180h = 1331, we get h = 1331/180 = 7.39 cm (to 2 decimal places). Therefore, the water level rises by 7.39 cm.
In simple words: When you put the cube in the water, it pushes the water up. The amount the water rises depends on how big the base is and how big the cube is.

Exam Tip: Remember that the volume of the submerged object equals the volume of water displaced. Keep the units consistent (all in cm or all in m) before calculating.

Question 15. A rectangular container, whose base is a square of side 6 cm, stands on a horizontal table and holds water up to 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 2 cm³ of water overflows. Calculate the volume of the cube.
Answer: The container base is a square with side 6 cm, so l = 6 cm and b = 6 cm. When the cube is submerged, water rises an additional 1 cm (to reach the top) and 2 cm³ spills over. By the principle of volume displacement: Volume of cube = Volume of water rise + Overflow volume = (6 × 6 × 1) + 2 = 36 + 2 = 38 cm³. Therefore, the volume of the cube is 38 cm³.
In simple words: The cube makes water rise by 1 cm and also causes 2 cubic centimetres to spill. Add these two amounts together to get the cube's volume.

Exam Tip: When liquid overflows, add the overflow volume to the height-rise calculation - both represent the displaced volume. The total displacement equals the volume of the submerged object.

Question 16. Two cubes, each with 12 cm edge, are joined end to end. Find the surface area of the resulting cuboid.
Answer: When two cubes of side 12 cm are joined end to end, they form a cuboid with dimensions: length = 12 + 12 = 24 cm, breadth = 12 cm, height = 12 cm. Using the total surface area formula for a cuboid: Surface area = 2(lb + bh + hl) = 2(24 × 12 + 12 × 12 + 12 × 24) = 2(288 + 144 + 288) = 2 × 720 = 1440 cm². Therefore, the surface area of the resulting cuboid is 1440 cm².
In simple words: When you join two cubes together, the length doubles but the width and height stay the same. Use the standard surface area formula with these new dimensions.

Exam Tip: Draw the joined figure to visualize which dimension changes. Always list the three dimensions of the resulting cuboid before applying the formula.

Question 17. A cube of a metal of 6 cm edge is melted and cast into a cuboid whose base is 9 cm × 8 cm. Find the height of the cuboid.
Answer: The cube has side 6 cm, so its volume is 6³ = 216 cm³. Since the same metal is melted and recast, the cuboid volume equals the cube volume. For the cuboid: Volume = l × b × h, so 216 = 9 × 8 × h. This gives h = 216/72 = 3 cm. Therefore, the height of the cuboid is 3 cm.
In simple words: When you melt and recast metal, the amount stays the same. Use the volume of the cube to find the missing dimension of the cuboid.

Exam Tip: In melting and casting problems, volumes are always equal. This is the key principle - set the two volumes equal and solve for the unknown dimension.

Question 18. The area of a playground is 4800 m². Find the cost of covering it with gravel 1 cm deep, if the gravel costs Rs. 260 per cubic metre.
Answer: The playground area is 4800 m². The depth of gravel is 1 cm, which equals 1/100 = 0.01 m. The volume of gravel needed is: Volume = Area × Depth = 4800 × 0.01 = 48 m³. At Rs. 260 per cubic metre, the total cost = 260 × 48 = Rs. 12,480. Therefore, the cost of covering the playground is Rs. 12,480.
In simple words: To find how much gravel you need, multiply the area by the depth. Then multiply that volume by the price per cubic metre.

Exam Tip: Always convert all measurements to the same unit before multiplying. Here, convert 1 cm to metres (0.01 m) to match the area unit of m².

Question 19. A field is 30 m long and 18 m broad. A pit 6 m long, 4 m wide and 3 m deep is dug out from the middle of the field and the earth removed is evenly spread over the remaining area of the field. Find the rise in the level of the remaining part of the field in centimeters correct to two decimal places.
Answer: The total field area is 30 × 18 = 540 m². The pit dimensions are 6 × 4 m, giving a pit area of 24 m². The remaining field area is 540 - 24 = 516 m². The earth dug out has volume: 6 × 4 × 3 = 72 m³. This earth is spread over the remaining area, so: 516 × h = 72, which gives h = 72/516 ≈ 0.1395 m = 13.95 cm. Therefore, the level of the remaining field rises by 13.95 cm.
In simple words: The soil from the pit is spread across the remaining ground. Divide the volume of dug soil by the remaining area to find how much higher the ground becomes.

Exam Tip: Remember to subtract the pit area from the total field area - only the remaining area gets the soil spread over it. This is a common mistake in these problems.

Question 20. A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?
Answer: The total plot area is 24 × 20 = 480 m². Each corner pit is a cube with side 4 m. The surface area of one cube is 4 × 4² = 64 m². Since there are 4 pits, the total pit area is 4 × 64 = 256 m². Wait - let me reconsider. The volume of each pit is 4³ = 64 m³, so total volume dug is 4 × 64 = 256 m³. The area of one square pit is 4 × 4 = 16 m², and for 4 pits that's 64 m². The remaining area is 480 - 64 = 416 m². Using Volume = Area × Height: 416h = 256, so h = 256/416 = 8/13 metre. Therefore, the remaining plot gets raised by 8/13 metre.
In simple words: Four cubes are dug from the corners. Find how much soil comes out, then spread that across the ground that remains after the pits are removed.

Exam Tip: When pits are at corners, each pit removes a square area from the total. Subtract all pit areas from the total, then divide the total excavated volume by this remaining area.

Question 21. The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m³ of wood costs Rs. 22,000.
Answer: The inner dimensions are 2 m, 1.2 m, and 0.75 m. Wood thickness is 2.5 cm = 0.025 m. The outer dimensions are: (2 + 2 × 0.025) = 2.05 m, (1.2 + 2 × 0.025) = 1.25 m, (0.75 + 2 × 0.025) = 0.80 m. The volume of wood used equals the difference between outer and inner volumes: Volume = (2.05 × 1.25 × 0.80) - (2 × 1.2 × 0.75) = 2.05 - 1.80 = 0.25 m³. The cost of wood is 22,000 × 0.25 = Rs. 5,500. Therefore, the cost of wood required is Rs. 5,500.
In simple words: Calculate the outer size by adding twice the thickness to each inner measurement. Subtract the inner volume from the outer volume to get the wood volume.

Exam Tip: When a box is closed, thickness is added on all sides, so multiply the thickness by 2 for each dimension. The wood forms a hollow shell - its volume is outer minus inner.

Question 22. A cubical wooden box of internal edge 1 m is made of 5 cm thick wood. The box is open at the top. If the wood costs Rs. 28,800 per cubic metre, find the cost of the wood required to make the box. Calculate the cost to nearest hundred rupees.
Answer: The inner edge is 1 m. Wood thickness is 5 cm = 0.05 m. Since the box is open at the top, the external dimensions are: External length = 1 + 2(0.05) = 1.1 m, External breadth = 1 + 2(0.05) = 1.1 m, External height = 1 + 0.05 = 1.05 m (only on the bottom, not the top). The volume of wood = Outer volume - Inner volume = (1.1 × 1.1 × 1.05) - (1 × 1 × 1) = 1.2705 - 1 = 0.2705 m³. The cost = 28,800 × 0.2705 = Rs. 7,790.40. Rounding to the nearest hundred, the cost is Rs. 7,800.
In simple words: Since the top is open, the height only adds thickness on one side. Calculate outer volume, subtract inner volume, then multiply by the cost per cubic metre.

Exam Tip: Pay careful attention to "open" conditions. An open top means no thickness is added to the height from above - only from the bottom. Always round currency to the nearest required unit (here, nearest hundred).

Question 23. A square brass plate of side x cm is 1 mm thick and weighs 4725 g. If one cc of brass weights 8.4 g, find the value of x.
Answer: We are given that the side of the square brass plate is x cm. Since it is square, both length and width equal x cm. The plate is 1 mm thick, which converts to 0.1 cm. Using the volume formula for a rectangular solid, volume equals length times width times height, so volume = x × x × 0.1 = 0.1x² cm³. Since 1 cm³ of brass weighs 8.4 g, the total weight of this plate is 8.4 × 0.1x² grams. We know the actual weight is 4725 g, so we set up the equation: 8.4 × 0.1x² = 4725. Simplifying: 0.84x² = 4725. Dividing both sides by 0.84 gives x² = 5625. Taking the square root: x = 75 cm.
In simple words: Find the volume by multiplying the three dimensions together. Then use the weight per unit volume to set up an equation and solve for x.

Exam Tip: Always convert units to the same system before calculating volume. Remember that the weight formula is volume × density per unit volume.

Question 24. Three cubes whose edges are x cm, 8 cm and 10 cm respectively are melted and recast into a single cube of edge 12 cm. Find x.
Answer: When the three cubes are melted and recombined into one cube, the total volume stays the same. The volume of a cube is the edge length cubed. So the sum of the three original volumes equals the volume of the new cube: x³ + 8³ + 10³ = 12³. Working out each cube: x³ + 512 + 1000 = 1728. Combining the numbers: x³ + 1512 = 1728. Solving for x³: x³ = 216. Since 216 = 6³, we get x = 6 cm.
In simple words: Add up the volumes of the three small cubes and set this equal to the volume of the big cube. Then find the unknown edge length.

Exam Tip: Always apply the principle that volume is conserved when melting and recasting solids - what you start with equals what you end with.

Question 25. The area of cross-section of a pipe is 3.5 cm² and water is flowing out of pipe at the rate of 40 cm/s. How much water is delivered by the pipe in one minute?
Answer: The water moves through the pipe at 40 cm per second, so in one second, a column of water 40 cm long passes through the cross-section. The volume that flows in 1 second equals the cross-sectional area multiplied by the length of this column: volume = 3.5 × 40 = 140 cm³ per second. In one minute there are 60 seconds, so the total volume delivered is 140 × 60 = 8400 cm³. Converting to litres using the fact that 1 litre = 1000 cm³: 8400 ÷ 1000 = 8.4 litres.
In simple words: Multiply the cross-sectional area by the distance water travels in one second to get the flow per second. Then multiply by 60 to get the flow for one minute.

Exam Tip: Remember to convert from cm³ to litres at the end using the conversion 1000 cm³ = 1 litre. Always note the time units in the question.

Question 26(a). The figure (i) given below shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in cm and all angles in the figure are right angles.
Answer: This solid can be divided into two cuboids by drawing a line through it. The first (vertical) cuboid has dimensions 4 × 2 × 6 cm, giving a volume of 48 cm³. The second (horizontal) cuboid has dimensions 4 × 4 × 2 cm, giving a volume of 32 cm³. The total volume of the solid is 48 + 32 = 80 cm³.
In simple words: Break the solid shape into two rectangular boxes, find the volume of each box separately, then add them together.

Exam Tip: When a solid has an unusual shape, try to split it into simpler shapes like rectangular boxes. Make sure your divisions do not overlap.

Question 26(b). The figure (ii) given below shows the cross section of a concrete wall to be constructed. It is 2 m wide at the top, 3.5 m wide at the bottom and its height is 6 m and its length is 400 m. Calculate (i) the cross sectional area and (ii) volume of concrete in the wall.
Answer:
(i) The cross-section forms a trapezium with parallel sides of 2 m (top) and 3.5 m (bottom). Using the trapezium area formula: Area = (1/2) × (sum of parallel sides) × height = (1/2) × (2 + 3.5) × 6 = (1/2) × 5.5 × 6 = 16.5 m².
(ii) The volume of concrete equals the cross-sectional area times the length of the wall: Volume = 16.5 × 400 = 6600 m³.
In simple words: The cross-section is a trapezium shape. Apply the trapezium formula to find its area. Then multiply this area by the length to get the total volume.

Exam Tip: Identify the shape of the cross-section first - it is crucial for choosing the right formula. For a trapezium, always identify which sides are parallel.

Question 26(c). The figure (iii) given below show the cross-section of a swimming pool 10 m broad, 2 m deep at one end and 3 m deep at the other end. Calculate the volume of water it will hold when full, given that its length is 40 m.
Answer: The cross-section of the pool is a trapezium with depths (parallel sides) of 2 m and 3 m, and the length between them is 40 m. First, find the cross-sectional area: Area = (1/2) × (2 + 3) × 40 = (1/2) × 5 × 40 = 100 m². The volume of water the pool holds equals this cross-sectional area times the width: Volume = 100 × 10 = 1000 m³.
In simple words: View the pool from the side to see a trapezium. Find its area using the trapezium formula. Multiply by the pool width to get the total volume.

Exam Tip: Be careful to identify which dimension is which - in this case, the trapezium cross-section uses the depths, and the width of the pool is multiplied separately.

Question 27. A swimming pool is 50 metres long and 15 metres wide. Its shallow and deep ends are 1.5 metres and 4.5 metres deep respectively. If the bottom of the pool slopes uniformly, find the amount of water required to fill the pool.
Answer: The pool's cross-section (viewed from the side along its length) forms a trapezium with parallel sides of 1.5 m and 4.5 m, separated by 50 m. Using the trapezium area formula: Area = (1/2) × (1.5 + 4.5) × 50 = (1/2) × 6 × 50 = 150 m². The total volume of water needed equals this cross-sectional area multiplied by the width of the pool: Volume = 150 × 15 = 2250 m³.
In simple words: Picture the pool from the side - you see a trapezium with the two depths as parallel sides. Calculate the trapezium's area, then multiply by the pool's width to get the volume.

Exam Tip: Always clarify the orientation of the shape in your mind. For sloped pools, the depth values become the parallel sides of a trapezium.

Question 1. Area of a triangle is 30 cm². If its base is 10 cm, then its height is
(1) 5 cm
(2) 6 cm
(3) 7 cm
(4) 8 cm
Answer: (2) 6 cm
In simple words: Use the triangle area formula with the known values to find height. Rearrange the formula to solve for the missing measurement.

Exam Tip: The triangle area formula is (1/2) × base × height - remember the factor of 1/2, as this is the most common mistake.

Question 2. If the perimeter of a square is 80 cm, then its area is
(1) 800 cm²
(2) 600 cm²
(3) 400 cm²
(4) 200 cm²
Answer: (3) 400 cm²
In simple words: Find the side length first using perimeter = 4 × side. Then calculate area using area = side × side.

Exam Tip: Break multi-step problems into stages - first find the side from perimeter, then use it to find area.

Question 3. Area of a parallelogram is 48 cm². If its height is 6 cm then its base is
(1) 8 cm
(2) 4 cm
(3) 16 cm
(4) None of these
Answer: (1) 8 cm
In simple words: The parallelogram area equals base times height. Divide the given area by the height to find the base.

Exam Tip: For parallelograms, the height must be measured perpendicular to the base, not along a slanted side.

Question 4. If d is the diameter of a circle, then its area is
(1) πd²
(2) πd²/2
(3) πd²/4
(4) 2πd²
Answer: (3) πd²/4
In simple words: The radius is half the diameter, so r = d/2. Substitute this into the standard area formula πr² to get the answer.

Exam Tip: When given diameter instead of radius, always convert by dividing by 2 before using the standard circle formulas.

Question 5. If the area of trapezium is 64 cm² and the distance between parallel sides is 8 cm, then sum of its parallel sides is
(1) 8 cm
(2) 4 cm
(3) 32 cm
(4) 16 cm
Answer: (4) 16 cm
In simple words: Rearrange the trapezium formula to find the sum of parallel sides. Divide the area by the distance and multiply by 2.

Exam Tip: The trapezium formula has 1/2 as a factor - when rearranging, multiply the area by 2 to clear this fraction.

Question 6. Area of a rhombus whose diagonals are 8 cm and 6 cm is
(1) 48 cm²
(2) 24 cm²
(3) 12 cm²
(4) 96 cm²
Answer: (2) 24 cm²
In simple words: For a rhombus, multiply the two diagonals and take half the result to get the area.

Exam Tip: The rhombus area formula uses diagonals, not sides - this makes it different from most other quadrilateral formulas.

Question 7. If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be
(1) doubled
(2) trebled
(3) four times
(4) halved
Answer: (3) four times
In simple words: When both diagonals are doubled, the area formula shows that (2d₁) × (2d₂) / 2 = 4 × (d₁ × d₂ / 2), so the area becomes 4 times larger.

Exam Tip: For area problems involving scaling, check how both dimensions change - squaring the scale factor often applies when both length and width scale the same way.

Question 7. When the diagonals of a rhombus are doubled, the area becomes
(1) doubled
(2) tripled
(3) four times
(4) remains same
Answer: (3) four times
In simple words: If you make both diagonals twice as long, the total space inside the rhombus becomes 4 times bigger.

Exam Tip: Remember that when a rhombus's diagonals scale by factor k, the area scales by k². For k = 2, area becomes 2² = 4 times larger.

Question 8. If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is
(1) 100 cm²
(2) 200 cm²
(3) 50 cm²
(4) None of these
Answer: (3) 50 cm²
In simple words: A diagonal splits the quadrilateral into two triangles. Add the areas of both triangles: one with height 4 cm and the other with height 6 cm, both using the 10 cm diagonal as their base.

Exam Tip: Always label the quadrilateral (ABCD), identify which diagonal divides it, and calculate the areas of the two triangles formed separately before adding them together.

Question 9. Area of a rhombus is 90 cm². If the length of one diagonal is 10 cm then the length of other diagonal is
(1) 18 cm
(2) 9 cm
(3) 36 cm
(4) 4.5 cm
Answer: (1) 18 cm
In simple words: Use the rhombus area formula with the known diagonal to find the missing one. Plug the numbers into the formula and solve for the unknown.

Exam Tip: Rearrange the area formula algebraically: if Area = ½ × d₁ × d₂, then d₂ = (2 × Area) / d₁. Substitute and calculate.

Question 10. In the adjoining figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is
(1) 11 cm
(2) 18 cm
(3) 25 cm
(4) 36 cm
Answer: (3) 25 cm
In simple words: A quadrant's perimeter includes one curved arc and two straight radii. Add the arc length to both radii lengths to get the total boundary.

Exam Tip: Perimeter of a quadrant = (πr/2) + 2r. This combines the quarter-circle arc with its two bounding radii.

Question 11. In the adjoining figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as center. The area of the shaded region is
(1) 10.5 cm²
(2) 38.5 cm²
(3) 49 cm²
(4) 11.5 cm²
Answer: (1) 10.5 cm²
In simple words: Find the square's area first. Then subtract the quadrant's area from it. The leftover space is what remains shaded.

Exam Tip: When a curved shape is inside a straight-sided shape, always calculate both areas separately, then subtract the unwanted part from the whole.

Question 12. The adjoining figure shows a rectangle and a semicircle. The perimeter of the shaded region is
(1) 70 cm
(2) 56 cm
(3) 78 cm
(4) 46 cm
Answer: (2) 56 cm
In simple words: The boundary of the shaded area includes the length and both widths of the rectangle, plus the curved semicircle arc. Add all these parts together.

Exam Tip: For composite perimeters, trace the boundary carefully - the side where the semicircle sits is not counted as a straight edge, only the arc is included.

Question 13. The area of the shaded region shown in the below figure is
(1) 140 cm²
(2) 77 cm²
(3) 294 cm²
(4) 217 cm²
Answer: (4) 217 cm²
In simple words: The shaded area consists of a rectangle and a semicircle sitting on top of it. Calculate each area and then add them.

Exam Tip: Always identify the component shapes (rectangle, semicircle, etc.), find each area separately, then combine them according to whether you are finding total area or difference.

Question 14. In the adjoining figure, the boundary of the shaded region consists of semicircular arcs. The area of the shaded region is equal to
(1) 616 cm²
(2) 385 cm²
(3) 231 cm²
(4) 308 cm²
Answer: (4) 308 cm²
In simple words: Even though the boundary is made up of multiple curved arcs, the total shaded space equals exactly half of the largest circle's area.

Exam Tip: When smaller semicircles are removed from a larger one, a clever geometry fact emerges - the remaining shaded area equals half the larger circle's full area, regardless of how many smaller curves were subtracted.

Question 15. The perimeter of the shaded region shown in the below figure is
(1) 44 cm
(2) 88 cm
(3) 66 cm
(4) 132 cm
Answer: (2) 88 cm
In simple words: The outer boundary is a large semicircular arc. Inside, two smaller semicircular arcs form part of the edge. Add the arc of the big semicircle to the arcs of both small semicircles.

Exam Tip: For perimeters involving semicircles, use πr for each semicircle arc. Add all arc lengths (not diameters) that actually form the boundary you are measuring.

Question 16. In the adjoining figure, ABC is a right angled triangle at B. A semicircle is drawn on AB as diameter. If AB = 12 cm and BC = 5 cm, then the area of the shaded region is
(1) (60 + 18π) cm²
(2) (30 + 36π) cm²
(3) (30 + 18π) cm²
(4) (30 + 9π) cm²
Answer: (3) (30 + 18π) cm²
In simple words: The shaded region includes both the triangle and the semicircle. Calculate the area of each shape and add them - the triangle contributes 30 cm², and the semicircle contributes 18π cm².

Exam Tip: When combining shapes, find each area using its own formula separately. For a right triangle, multiply half the base by the height; for a semicircle, use πr²/2.

Question 17. The perimeter of the shaded region shown in the below figure is
(1) (30 + 6π) cm
(2) (30 + 12π) cm
(3) (18 + 12π) cm
(4) (18 + 6π) cm
Answer: (2) (30 + 12π) cm
In simple words: The boundary consists of two straight sides of the right triangle (the two legs that are not the hypotenuse) plus the curved semicircular arc. Add 12 + 5 + 13 for the three sides, but replace the hypotenuse (13 cm) with the semicircle's arc measurement to find the actual perimeter.

Exam Tip: For composite perimeters, carefully identify which edges form the actual boundary - the hypotenuse is replaced by the semicircle arc, so do not count it as a straight edge.

Question 17. In right angle triangle ABC, with sides AB = 12 cm and BC = 5 cm, a semi-circle is drawn on AB as diameter. Find the perimeter of the shaded region.
Answer: Using the Pythagorean theorem in the right triangle, we can work out the hypotenuse AC. Since \( AC^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169 \), we get \( AC = 13 \) cm. The semi-circle has a diameter of AB, so its radius is \( r = \frac{12}{2} = 6 \) cm. The perimeter includes three parts: the hypotenuse AC, the base BC, and the curved part of the semi-circle. The perimeter of the shaded region is \( AC + CB + \pi r = 13 + 5 + 6\pi = (18 + 6\pi) \) cm.
In simple words: First find the longest side AC using the Pythagorean rule. Then add up the three edges that form the border of the shaded region - the two straight sides and the curved semi-circle.

Exam Tip: Always identify which parts form the perimeter boundary - do not include interior lines. Remember to use \( \pi r \) for the semi-circle arc, not the full circumference \( 2\pi r \).

Question 18. If the volume of a cube is 729 m³, then its surface area is
(a) 486 cm²
(b) 324 cm²
(c) 162 cm²
(d) None of the options
Answer: (a) 486 cm²
In simple words: Find the side length by taking the cube root of the volume. Once you know the side, multiply the area of one face (side²) by 6 to get the total surface area.

Exam Tip: Surface area of a cube uses the formula \( 6s^2 \), where s is the side length. Always cube root the volume first to find s, then apply the surface area formula.

Question 19. If the total surface area of a cube is 96 cm², then the volume of cube is
(a) 8 cm³
(b) 512 cm³
(c) 64 cm³
(d) 27 cm³
Answer: (c) 64 cm³
In simple words: From the surface area, work backwards to find the side length. Use the formula \( 6s^2 = 96 \) to get s, then cube it to find the volume.

Exam Tip: When surface area is given, solve for the side using \( s = \sqrt{\frac{\text{Surface Area}}{6}} \), then calculate volume using \( s^3 \).

Question 20. The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
Answer: (a) 15 m
In simple words: The longest pole that fits in a box-shaped room is the diagonal that goes from one corner to the opposite corner. Use the 3D distance formula with the three dimensions.

Exam Tip: The diagonal of a cuboid is \( \sqrt{l^2 + b^2 + h^2} \). This represents the longest straight line that can fit inside the rectangular space.

Question 21. The lateral surface area of a cube is 256 m². The volume of the cube is
(a) 512 m³
(b) 64 m³
(c) 216 m³
(d) 256 m³
Answer: (a) 512 m³
In simple words: Lateral surface area means only the four vertical sides, not the top and bottom. Use \( 4s^2 \) to find the side, then cube it for the volume.

Exam Tip: Remember that lateral (side) surface area uses 4 faces only (\( 4s^2 \)), while total surface area uses all 6 faces (\( 6s^2 \)). Do not confuse these two formulas.

Question 22. If the perimeter of one face of a cube is 40 cm, then the sum of lengths of its edges is
(a) 80 cm
(b) 120 cm
(c) 160 cm
(d) 240 cm
Answer: (b) 120 cm
In simple words: Each face is a square. From the perimeter of one square face, find the side length. A cube has 12 edges total, so multiply the side length by 12.

Exam Tip: A cube has 12 edges in total. Once you find the side from the perimeter (divide by 4), multiply by 12 to get the sum of all edge lengths.

Question 23. A cuboid container has the capacity to hold 50 small boxes. If all the dimensions of the container are doubled, then it can hold (small boxes of same size)
(a) 100 boxes
(b) 200 boxes
(c) 400 boxes
(d) 800 boxes
Answer: (c) 400 boxes
In simple words: When all three dimensions are doubled, the volume increases by a factor of \( 2 \times 2 \times 2 = 8 \). So the container can now hold 8 times as many boxes.

Exam Tip: If dimensions are scaled by factor k, volume scales by \( k^3 \). For k = 2, the volume becomes 8 times larger, so capacity multiplies by 8.

Question 24. The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is
(a) 1900
(b) 1920
(c) 1800
(d) 1840
Answer: (b) 1920
In simple words: Convert all measurements to the same unit. Calculate the volume of one plank and the volume of the pit. Divide the pit volume by the plank volume to find how many fit inside.

Exam Tip: Always convert smaller units (cm) to larger units (m) before calculating volume. The number of objects that fit equals total volume divided by individual volume.

Question 25. Consider the following two statements:
Statement 1: If the circumference of a circle is 10π cm, then its area is 25π cm².
Statement 2: The area of a circle is π times its circumference.
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (c) Statement 1 is true, and Statement 2 is false.
In simple words: For Statement 1, use the circumference to find the radius, then check if the area matches. For Statement 2, compare the formulas: area is \( \pi r^2 \) and circumference is \( 2\pi r \) - these are not simply related by a factor of π.

Exam Tip: Always verify each statement independently by working through the math. Do not assume statements are connected - check their validity separately using the definitions of circumference and area.

Question. Assertion (A): Sides of a triangle are 9 cm, 12 cm and 15 cm. This triangle is both scalene triangle and right angle triangle.
Reason (R): Area of a right triangle = \( \frac{1}{2} \times \) base × height.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
In simple words: Check if the three sides form a scalene triangle (all different) and a right triangle (using Pythagorean theorem). The reason given is a true formula, but it does not explain why the triangle is both scalene and right-angled.

Exam Tip: In Assertion-Reason questions, both parts can be true independently, but the reason must actually explain the assertion. A correct formula is not the same as a correct explanation for a specific claim.

Question. Assertion (A): Heron's formula can be used to find the area of a scalene triangle only.
Reason (R): If ABC is a triangle with side a, b and c respectively, then its area = \( \sqrt{s(s - a)(s - b)(s - c)} \), where \( s = \frac{a + b + c}{2} \).
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true.
In simple words: Heron's formula works for any type of triangle - scalene, isosceles, or equilateral. The reason is a correct statement of the formula itself, but the assertion is wrong because it says the formula only works for scalene triangles.

Exam Tip: Heron's formula is a universal method for finding the area of any triangle when all three sides are known. Do not limit its application to one specific type of triangle.

Question. Assertion (A): The volume of a cuboid having length, breadth and diagonal as 4 m, 3 m and 13 m is 144 m³.
Reason (R): Length of diagonal of a cuboid is \( \sqrt{l^2 + b^2 + h^2} \).
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: Use the diagonal formula with the known dimensions to find the missing height. The reason directly helps verify the assertion by allowing us to calculate the third dimension.

Exam Tip: In this case, the reason is the exact mathematical tool needed to solve the assertion. The diagonal formula allows you to find the missing dimension, then calculate the volume.

Chapter Test

Question 1(a). Calculate the area of the shaded region.
Answer: [Figure showing a composite shape required - unable to complete answer without figure context]
In simple words: Identify all shapes that make up the shaded region. Calculate the area of each shape separately, then add or subtract as needed depending on whether areas overlap or are adjacent.

Exam Tip: For composite shaded regions, break them into simpler shapes (rectangles, circles, triangles). Calculate each area using appropriate formulas, then combine them correctly using addition or subtraction.

Question 1(a). Find the area of the shaded region shown in the figure. All measurements are in centimeters.
Answer: Starting with triangle AOB, which has a right angle at O, we can work out its area using the base and height method. The area comes to 30 cm². Next, we use the Pythagorean theorem to find the length AB. Since AO is 12 cm and OB is 5 cm, we get AB = 13 cm. For triangle ABC with sides of 13, 14, and 15 cm, we apply Heron's formula. First, we calculate the semi-perimeter: s = 21 cm. Substituting into Heron's formula, the area of triangle ABC is 84 cm². The shaded region is found by subtracting the area of triangle AOB from the area of triangle ABC: 84 - 30 = 54 cm².
In simple words: Find the area of the smaller triangle using base times height. Then find the area of the bigger triangle using Heron's formula. Subtract the smaller from the bigger to get the shaded area.

Exam Tip: Always check that you have correctly identified which triangle is the shaded region and which are the unshaded parts. Remember to use Heron's formula when you have three different side lengths.

Question 1(b). If the sides of a square are lengthened by 3 cm, the area becomes 121 cm². Find the perimeter of the original square.
Answer: Let the original side length be x cm. When the sides are lengthened by 3 cm, the new side becomes (x + 3) cm. According to the problem, the area of the new square is 121 cm², which gives us the equation (x + 3)² = 121. Expanding: x² + 6x + 9 = 121, which simplifies to x² + 6x - 112 = 0. Factoring this quadratic: (x - 8)(x + 14) = 0. This gives x = 8 or x = -14. Since length cannot be negative, x = 8 cm. The perimeter of the original square is 4 × 8 = 32 cm.
In simple words: Set up an equation using the new area. Solve the quadratic by factoring. Reject the negative answer because a side length cannot be negative. Multiply by 4 to get the perimeter.

Exam Tip: When solving quadratic equations from word problems, always reject negative solutions since measurements must be positive. Show your factorization clearly.

Question 2(a). Find the area enclosed by the figure (i) given below. All measurements are in centimeters.
Answer: First, we determine that the outer boundary forms a square ABCD with side 9 cm. Inside, triangles EFG and HIJ are congruent by the SAS criterion (they have equal bases and heights, and both have right angles). Each triangle has an area of 15 cm². The area of square ABCD is 81 cm². The shaded region is calculated by subtracting both triangle areas from the square's area: 81 - 15 - 15 = 51 cm².
In simple words: Find the area of the big square. Find the area of each small triangle. Subtract both triangle areas from the square to get the shaded part.

Exam Tip: Always check congruence criteria carefully when comparing triangles - this helps you avoid calculating the same triangle area twice. Verify your subtraction is correct.

Question 2(b). Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimeters.
Answer: The quadrilateral ABCD is divided into two triangles by diagonal BD. In right triangle ABD, we apply the Pythagorean theorem: BD² = 6² + 8² = 100, so BD = 10 cm. In right triangle BDC, we have BC² = BD² + DC², which gives 26² = 10² + DC², so DC = 24 cm. The area of triangle ABD is calculated as half the product of its two perpendicular sides: (1/2) × 8 × 6 = 24 cm². The area of triangle BDC is: (1/2) × 24 × 10 = 120 cm². Adding these together: 24 + 120 = 144 cm².
In simple words: Draw the diagonal to split the quadrilateral into two right triangles. Use the Pythagorean theorem to find missing sides. Find each triangle's area and add them together.

Exam Tip: Identify the right angles in your triangles before applying the Pythagorean theorem. Ensure you correctly identify which sides are the base and height for area calculations.

Question 2(c). Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in meters.
Answer: The outer square ABCD has a side of 12 m, giving it an area of 144 m². We then identify and calculate the areas of the unshaded regions. Triangle AEH (a right triangle) has an area of 12.5 m². Triangle EBF has an area of 17.5 m². Trapezium GDCF has parallel sides of 3 m and 7 m with a height of 12 m, giving an area of 60 m². The total unshaded area is 12.5 + 17.5 + 60 = 90 m². The shaded area is found by subtracting: 144 - 90 = 54 m².
In simple words: Find the area of the big square. Find the areas of all the white shapes inside. Subtract all the white areas from the square's area to get the gray shaded part.

Exam Tip: Carefully identify each unshaded region - some are triangles and some are trapeziums. Use the correct formula for each shape and add all unshaded areas before subtracting from the total.

Question 3. Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.
Answer: The aeroplane shape is divided into five component parts. For triangle ABK with sides 1 cm, 5 cm, and 5 cm, we apply Heron's formula. The semi-perimeter is 5.5 cm, and the area works out to 2.5 cm². Triangle KIJ (a right triangle) has dimensions 1.5 cm and 6 cm, giving an area of 4.5 cm². Triangle CBD also measures 4.5 cm². Rectangle BKHE has length 6.5 m and width 1 cm, giving 6.5 cm². For the trapezium EFGH, since it is isosceles with parallel sides of 1 cm and 2 cm, the perpendicular distance is found using the Pythagorean theorem to be 0.87 cm, making the trapezium's area 1.3 cm². Adding all parts: 2.5 + 4.5 + 4.5 + 6.5 + 1.3 = 19.3 cm².
In simple words: Break the aeroplane shape into simpler pieces like triangles, rectangles, and trapeziums. Find the area of each piece using the right formula. Add all the pieces together to get the total.

Exam Tip: When dealing with composite shapes, always identify all distinct regions clearly and use Heron's formula when side lengths are given but it is not a right triangle. Check your final answer is correct to one decimal place as instructed.

Question 4. If the area of a circle is 78.5 cm², find its circumference. (Take π = 3.14)
Answer: We begin by using the area formula for a circle, πr² = 78.5 cm². With π = 3.14, we can solve for the radius: r² = 78.5 ÷ 3.14 = 25, so r = 5 cm. Once we have the radius, we use the circumference formula C = 2πr = 2 × 3.14 × 5 = 31.4 cm.
In simple words: Use the area to find the radius. Once you know the radius, multiply it by 2π to get the circumference.

Exam Tip: Always use the given value of π consistently throughout your calculation. Show both the radius calculation and the circumference calculation as separate steps for full marks.

Question 5. From a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm², calculate the original area of the cardboard.
Answer: Let the radius of the circle be r cm. We know the area of the circle is 154 cm², so \( \frac{22}{7} \times r^2 = 154 \) which gives us \( r^2 = \frac{154 \times 7}{22} = 49 \), therefore \( r = 7 \) cm. Since the biggest circle that can fit inside a square has its diameter equal to the side of the square, the side of the square is \( 2r = 2 \times 7 = 14 \) cm. The area of the cardboard is \( (side)^2 = 14^2 = 196 \) cm².
In simple words: The circle's radius is 7 cm, so the square's side is 14 cm, making the cardboard's area 196 cm².

Exam Tip: Remember that the largest circle fitting in a square has its diameter equal to the square's side - this relationship is key to solving this type of problem quickly.

Question 6(a). From a sheet of paper of dimensions 2 m × 1.5 m, how many circles of radius 5 cm can be cut? Also find the area of the paper wasted. (Take π = 3.14)
Answer: Converting the sheet dimensions to cm: length = 200 cm and breadth = 150 cm. Each circle has radius 5 cm, so diameter = 10 cm. The number of circles that can be cut lengthwise is \( \frac{200}{10} = 20 \), and breadthwise is \( \frac{150}{10} = 15 \). Therefore, total circles = \( 20 \times 15 = 300 \). The area of each circle is \( 3.14 \times 5^2 = 78.5 \) cm². Total area of 300 circles = \( 300 \times 78.5 = 23550 \) cm². The area of paper wasted = Area of sheet - Area of circles = \( 200 \times 150 - 23550 = 30000 - 23550 = 6450 \) cm².
In simple words: You can arrange 300 circles (20 by 15) on the sheet, and the leftover paper covers 6450 cm².

Exam Tip: Always convert units first to keep calculations consistent, and remember that wasted area equals total sheet area minus the sum of all cut-out shapes.

Question 6(b). If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.
Answer: The semi-circular protractor has diameter 14 cm, so radius \( r = \frac{14}{2} = 7 \) cm. The perimeter of a semi-circle consists of the curved part (half the circumference) plus the diameter. Perimeter = \( \pi r + 2r = \frac{22}{7} \times 7 + 2 \times 7 = 22 + 14 = 36 \) cm.
In simple words: The perimeter includes the curved half-circle arc and the straight diameter at the base, which sum to 36 cm.

Exam Tip: A semi-circle's perimeter has two parts - the curved arc and the straight base - don't forget either one or you'll lose marks.

Question 7. A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of Rs 60 per square metre.
Answer: The circumference of the circular park is 88 m, so \( 2\pi r = 88 \), which gives \( 2 \times \frac{22}{7} \times r = 88 \), therefore \( r = 14 \) m. The road is 3.5 m wide, so the outer radius is \( R = 14 + 3.5 = 17.5 \) m. The area of the road is the difference between the outer and inner circles: \( \pi R^2 - \pi r^2 = \pi[(17.5)^2 - (14)^2] = \frac{22}{7}[306.25 - 196] = \frac{22}{7} \times 110.25 = 346.5 \) m². The total cost is \( 346.5 \times 60 = \) Rs 20790.
In simple words: The road's area is found by subtracting the inner park's area from the outer circle's area, then multiply by the cost per square metre.

Exam Tip: For ring-shaped areas, always calculate the outer circle area minus the inner circle area - this avoids errors from trying to compute the area directly.

Question 8. The adjoining sketch shows a running track 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.
Answer: From the figure, the inner rectangle ABCD has length 140 m and breadth 42 m. The inner semi-circles have diameter 42 m, so radius \( r = 21 \) m. The outer rectangle has the same length (140 m) and breadth \( 42 + 2(3.5) = 49 \) m. The outer semi-circles have radius \( R = \frac{49}{2} = 24.5 \) m. The area of the track equals the area of the outer rectangle minus the inner rectangle, plus twice the difference of the outer and inner semi-circle areas: \( 140(49 - 42) + 2 \times \pi \times \frac{1}{2}(R^2 - r^2) = 140 \times 7 + \frac{22}{7}[(24.5)^2 - (21)^2] = 980 + \frac{22}{7} \times 159.25 = 980 + 500.5 = 1480.5 \) m².
In simple words: The track area is the outer shape's area minus the inner shape's area, combining rectangular and semi-circular parts.

Exam Tip: Break complex shapes into simpler parts (rectangles and semi-circles), calculate each part separately, then combine them - this method reduces mistakes.

Question 9. In the adjoining figure, O is the center of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)
Answer: Since the angle in a semi-circle is 90°, we have \( \angle ACB = 90° \). Using the Pythagorean theorem on triangle ABC with AC = 12 cm and BC = 16 cm: \( AB^2 = 12^2 + 16^2 = 144 + 256 = 400 \), so \( AB = 20 \) cm. Since AB is the diameter, the radius is \( r = 10 \) cm. The perimeter of the shaded region consists of the semi-circle arc plus the two sides AC and CB: Perimeter = \( \pi r + AC + BC = 3.142 \times 10 + 12 + 16 = 31.42 + 28 = 59.42 \) cm. The area of the shaded region is the semi-circle area minus the triangle area: \( \frac{\pi r^2}{2} - \frac{1}{2} \times AC \times BC = \frac{3.142 \times 100}{2} - \frac{1}{2} \times 12 \times 16 = 157.1 - 96 = 61.1 \) cm².
In simple words: The shaded shape is made from a semi-circle with a triangle removed from inside it.

Exam Tip: For shapes combining curves and straight lines, calculate perimeter and area separately - include all boundary parts (arcs and line segments) in the perimeter.

Question 10(a). In the figure (i) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle.
Answer: The perimeter of a quarter circle includes one quarter of the circumference plus two radii. Perimeter = \( \frac{2\pi r}{4} + 2r = \frac{\pi r}{2} + 2r \). Substituting \( r = 3.5 \) cm and \( \pi = \frac{22}{7} \): \( \frac{\frac{22}{7} \times 3.5}{2} + 2 \times 3.5 = \frac{22 \times 0.5}{2} + 7 = \frac{11}{2} + 7 = 5.5 + 7 = 12.5 \) cm.
In simple words: A quarter circle's perimeter adds one quarter of the full circumference to two straight radii.

Exam Tip: Always remember that the perimeter of a quarter-circle includes both the curved arc and two straight radii - a common error is forgetting the radii.

Question 10(b). In the figure (ii) given below, there are five squares each of side 2 cm.
(i) Find the radius of the circle.
(ii) Find the area of the shaded region. (Take π = 3.14).
Answer:
(i) Let O be the center of the circle and B be the midpoint of one side of the squares. From the figure, \( OB = 2 + 1 = 3 \) cm (since B is 1 cm from the center horizontally and 2 cm down from the top), and \( AB = 1 \) cm. Using the Pythagorean theorem: \( OA = \sqrt{OB^2 + AB^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \) cm. Therefore, the radius of the circle is \( \sqrt{10} \) cm.
(ii) The area of the circle is \( \pi r^2 = 3.14 \times (\sqrt{10})^2 = 3.14 \times 10 = 31.4 \) cm². The area of the five squares is \( 5 \times 2^2 = 5 \times 4 = 20 \) cm². The shaded area is \( 31.4 - 20 = 11.4 \) cm².
In simple words: The circle encloses some squares. The shaded part is the circle's area with the squares subtracted from it.

Exam Tip: When a shape is partially inside and outside a circle, find the total circle area first, then subtract or add the non-circular parts to get the shaded region.

Question 11(a). In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by the perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.
Answer: The area of the quadrant is \( \frac{\pi r^2}{4} = \frac{\frac{22}{7} \times 7^2}{4} = \frac{\frac{22}{7} \times 49}{4} = \frac{22 \times 7}{4} = \frac{154}{4} = 38.5 \) cm². The area of triangle OAB is \( \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 3 \times 4 = 6 \) cm². The area of the remaining piece is \( 38.5 - 6 = 32.5 \) cm². For the perimeter, we need the arc AYB (which is a quarter circle arc of radius 7), plus the line segments AB, BX, and XA. The arc length is \( \frac{2\pi r}{4} = \frac{\pi \times 7}{2} = \frac{\frac{22}{7} \times 7}{2} = 11 \) cm. Using the Pythagorean theorem on triangle OAB: \( AB = \sqrt{OA^2 + OB^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \) cm. The remaining segments are \( XA = 7 - 3 = 4 \) cm and \( YB = 7 - 4 = 3 \) cm. Perimeter = \( 11 + 5 + 4 + 3 = 23 \) cm.
In simple words: Remove a triangle from a quarter-circle slice. The remaining area is the quadrant minus the triangle; the perimeter includes the arc plus three straight sides.

Exam Tip: When a section is removed from a shape, carefully identify which boundary segments remain - the perimeter should include only the outer edges of what's left, not internal cuts.

Question 11(b). In the figure (ii) given below, ABCD is a square. Points A, B, C and D are centres of quadrants of circles of the same radius. If the area of the shaded portion is 21 3/7 cm², find the radius of the quadrants.
Answer: Let the radius of each quadrant be r cm. The area of one quadrant is πr²/4. So the combined area of all 4 quadrants is πr². Since the side of the square equals r + r = 2r, the square's area is (2r)² = 4r². The shaded area is found by subtracting the total quadrant area from the square's area: 4r² - πr² = 21 3/7. Converting the mixed number: 21 3/7 = 150/7. So we get: 4r² - πr² = 150/7, which simplifies to r²(4 - π) = 150/7. Substituting π = 22/7: r²(4 - 22/7) = 150/7, giving r²(28/7 - 22/7) = 150/7, so r²(6/7) = 150/7. Multiplying both sides by 7/6: r² = 150/6 = 25, therefore r = 5 cm.
In simple words: The shaded region is what remains when you remove four quarter-circles from a square. Use the given shaded area to set up an equation, solve for r, and you get 5 cm.

Exam Tip: Always express the given area as a single fraction first. Key is recognizing that the side of the square = 2r (sum of two radii from opposite corners).

Question 12. In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.
Answer: Since ABC is a right-angled triangle, by the Pythagorean theorem: AC² = AB² + BC² ...(1). The area of a semicircle with diameter AB is: π(AB/2)²/2 = πAB²/8 ...(2). The area of a semicircle with diameter BC is: π(BC/2)²/2 = πBC²/8 ...(3). The area of a semicircle with diameter AC is: π(AC/2)²/2 = πAC²/8 ...(4). Adding equations (2) and (3): πAB²/8 + πBC²/8 = π(AB² + BC²)/8. From equation (1), AB² + BC² = AC², so: π(AB² + BC²)/8 = πAC²/8, which matches equation (4). Therefore, the sum of the semicircle areas on AB and BC equals the semicircle area on AC.
In simple words: The Pythagorean theorem lets us show that the two smaller semicircles together have the same area as the large semicircle on the hypotenuse.

Exam Tip: The key insight is applying the Pythagorean relation to the areas. Always show the three area formulas and the final comparison explicitly for full marks.

Question 13. The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.
Answer: The minute hand sweeps out a sector of a circle. In 15 minutes, the hand moves through an angle of (15/60) × 360° = 90°. Using the sector area formula: Area = πr² × (θ/360°), where r = 14 cm and θ = 90°. Substituting: Area = (22/7) × 14² × (90/360) = (22/7) × 196 × (1/4) = 22 × 28 × (1/4) = 22 × 7 = 154 cm².
In simple words: In 15 minutes, the minute hand rotates through one quarter of a full circle (90°). Calculate a quarter of the full circle's area using the sector formula to get 154 cm².

Exam Tip: Always convert minutes to degrees first by using the ratio (minutes/60) × 360°. Then apply the sector area formula directly.

Question 14. Find the radius of a circle if a 90° arc has a length of 3.5π cm. Hence, find the area of the sector formed by this arc.
Answer: For a sector with central angle θ and radius r, the arc length is given by: L = (θ/360°) × 2πr. Here, θ = 90° and L = 3.5π. So: 3.5π = (90/360) × 2πr = (1/4) × 2πr = πr/2. Solving for r: r/2 = 3.5, thus r = 7 cm. Now, the sector area is: A = πr² × (θ/360°) = (22/7) × 7² × (90/360) = (22/7) × 49 × (1/4) = 22 × 7 × (1/4) = 154/4 = 38.5 cm².
In simple words: Use the arc length formula to find r = 7 cm first. Then use the sector area formula with this radius and the 90° angle to get 38.5 cm².

Exam Tip: Solve for the radius using the arc length equation before calculating the sector area. Both formulas involve the angle in degrees over 360°.

Question 15. A cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.
Answer: For a cube with edge length 28 cm, the maximum circle that fits on each face has diameter equal to the edge length. Thus, the radius is 28/2 = 14 cm. Each face is a 28 × 28 square. The unpainted area on each face equals the square's area minus the circle's area: 28² - π(14)² = 784 - (22/7) × 196 = 784 - 616 = 168 cm². Since the cube has 6 faces, the total unpainted surface area is: 6 × 168 = 1008 cm².
In simple words: Each face has a painted circle with radius 14 cm. The leftover space on each face is 168 cm². Multiply by 6 faces to get the total unpainted area of 1008 cm².

Exam Tip: The maximum circle on a square face has its diameter equal to the side of the square. Always subtract the circle area from the square area for each face, then multiply by the number of faces.

Question 16. Can a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5 m, 3 m and 4 m?
Answer: The longest straight object that can fit inside a cuboid is along its interior diagonal. The diagonal of a cuboid with dimensions l, b, h is: d = √(l² + b² + h²). Here: d = √(3.5² + 3² + 4²) = √(12.25 + 9 + 16) = √37.25 ≈ 6.12 m. Since the pole is 6.5 m long and the diagonal is only about 6.12 m, the pole cannot fit into the truck body.
In simple words: The longest straight pole that fits is along the diagonal of the truck's interior box. The diagonal is about 6.12 m, which is less than 6.5 m, so the pole does not fit.

Exam Tip: Always use the space diagonal formula, not just the length or width. The diagonal gives the maximum length an object can have inside the cuboid.

Question 17. A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol?
Answer: First, find the tank's volume: V = l × b × h = 40 × 28 × 25 = 28,000 cm³. Convert to litres using 1 litre = 1,000 cm³: 28,000 cm³ = 28 litres. Since the car travels 13.5 km per litre, the total distance for a full tank is: 28 × 13.5 = 378 km.
In simple words: Calculate the tank volume in cubic centimetres, convert to litres, then multiply by the fuel efficiency in km per litre to get the total distance of 378 km.

Exam Tip: Remember the conversion: 1 litre = 1,000 cm³. Always convert volume units before multiplying by the consumption rate.

Question 18. An aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.
Answer: Let the height be h cm. The aquarium's volume is: V = l × b × h = 200 cm × 80 cm × h cm = 16,000h cm³ = 16h litres. The water filling rate is 25 litres per 2 minutes = 12.5 litres per minute. In 96 minutes, the water filled is: 96 × 12.5 = 1,200 litres. Since the aquarium is now full: 16h = 1,200, so h = 1,200 ÷ 16 = 75 cm.
In simple words: Find the total water that filled the tank in 96 minutes (1,200 litres). This equals the tank's volume in litres (16h). Solve to get h = 75 cm.

Exam Tip: Convert all measurements to the same unit (cm) at the start, then convert the final volume to litres. Set this equal to the total water filled to solve for h.

Question 19. The lateral surface area of a cuboid is 224 cm². Its height is 7 cm and the base is a square. Find (i) a side of the square, and (ii) the volume of the cuboid.
Answer: (i) Since the base is a square, let each side be x cm. The lateral surface area formula is: LSA = 2(l + b) × h. Here: 224 = 2(x + x) × 7 = 2(2x) × 7 = 28x. Solving: x = 224 ÷ 28 = 8 cm. (ii) The volume is: V = l × b × h = 8 × 8 × 7 = 448 cm³.
In simple words: (i) Use the lateral surface area formula with l = b = x to find x = 8 cm. (ii) Then calculate the volume using these dimensions: 448 cm³.

Exam Tip: For a cuboid with a square base, both length and breadth equal x. The lateral surface area excludes the top and bottom, so use the formula 2(l + b) × h carefully.

Question 20. If the volume of a cube is V m³, its surface area is S m² and the length of a diagonal is d metres, prove that 6√3 V = Sd.
Answer: Let the side of the cube be a m. Then: Volume V = a³, Surface area S = 6a², and the space diagonal d = a√3. Now calculate 6√3 V: 6√3 V = 6√3 × a³ = 6√3a³. Calculate Sd: Sd = 6a² × a√3 = 6√3a³. Since both equal 6√3a³, we have 6√3 V = Sd.
In simple words: Express V, S, and d in terms of the side length a. Then compute both 6√3 V and Sd and show they are equal.

Exam Tip: Remember that the space diagonal of a cube with side a is a√3, not a√2. Substitute carefully and verify both sides simplify to the same expression.

Question 21. The adjoining figure shows a victory stand, each face is rectangular. All measurements are in centimetres. Find its volume and surface area (the bottom of the stand is open).
Answer: The stand is made of three rectangular parts. Volume of part (3): 50 × 40 × 12 = 24,000 cm³. Volume of part (1): 50 × 40 × (16 + 24) = 50 × 40 × 40 = 80,000 cm³. Volume of part (2): 50 × 40 × 24 = 48,000 cm³. Total volume: 24,000 + 80,000 + 48,000 = 152,000 cm³. For the surface area, calculate the exposed faces of each part (since the bottom is open and internal surfaces between parts are not counted). The total surface area is the sum of all outer rectangular faces that are visible, accounting for the openings and overlaps where parts meet. Based on the stand's structure, the surface area works out to be the combined lateral and top faces of all three parts.
In simple words: Break the stand into three boxes. Calculate each volume separately, then add them. For surface area, count only the faces that are visible from outside, excluding the bottom and any internal overlapping faces.

Exam Tip: Always decompose a composite solid into simpler shapes. For surface area, be careful to exclude the open bottom and any internal surfaces where parts connect. Sketch or visualize which faces are actually exposed.

Question 22. The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find
(i) the capacity of the box
(ii) the volume of the wood used in making the box, and
(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm³ of wood weighs 0.8 g.
Answer: Since the box is open and the wood is 2 cm thick everywhere, we first work out the internal dimensions. The external box measures 98 cm by 84 cm by 77 cm. Because one side is open (top), we subtract 2 cm from the thickness on only that side. On the other two sides (length and width), we subtract 4 cm total (2 cm from each face). This gives internal dimensions of 94 cm by 80 cm by 75 cm.

(i) The capacity (or inner volume) equals 94 × 80 × 75 = 564000 cm³.

(ii) To find the wood volume, we calculate the external volume minus the internal volume. External volume = 98 × 84 × 77 = 633864 cm³. Wood volume = 633864 - 564000 = 69864 cm³.

(iii) Since each cubic centimetre of wood has a mass of 0.8 g, the total mass is 69864 × 0.8 = 55891.2 g. Converting to kilograms: 55891.2 ÷ 1000 = 55.89 kg, which rounds to 55.9 kg.
In simple words: Find what fits inside the box (capacity). Then find how much wood makes up the box walls (subtract inside from outside). Finally, multiply the wood volume by 0.8 to get the weight, and convert grams to kilograms.

Exam Tip: Remember that an open box loses the top, so thickness applies differently — subtract 2 cm from height but 4 cm from length and width. Always verify your volume subtractions before converting units.

Question 23. A cuboidal block of metal has dimensions 36 cm by 32 cm by 0.25 m. It is melted and recast into cubes with an edge of 4 cm.
(i) How many such cubes can be made?
(ii) What is the cost of silver coating the surfaces of the cubes at the rate of Rs.1.25 per square centimeter?
Answer:
(i) First, convert all dimensions to the same unit. The cuboidal block is 36 cm by 32 cm by 0.25 m. Converting 0.25 m to centimetres: 0.25 × 100 = 25 cm. The volume of the block is 36 × 32 × 25 = 28800 cm³. Each small cube has edges of 4 cm, so its volume is 4 × 4 × 4 = 64 cm³. The number of cubes = 28800 ÷ 64 = 450 cubes.

(ii) Each cube has a total surface area of 6 × (side)² = 6 × 4² = 6 × 16 = 96 cm². For 450 cubes, the combined surface area is 450 × 96 = 43200 cm². At Rs.1.25 per cm², the total cost = 43200 × 1.25 = Rs.54000.
In simple words: Divide the big block's volume by one small cube's volume to find how many cubes you get. Then multiply the number of cubes by the surface area of one cube, and multiply that total surface area by the cost per square centimetre.

Exam Tip: Always convert units before calculating volume — mixing metres and centimetres is a common source of error. Double-check your surface area formula: 6 × (edge)², not 4 × (edge)².

Question 24. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of Rs.3.50 per square centimeter.
Answer: We calculate the volume of each small cube and add them to find the total volume. The first cube has volume 3³ = 27 cm³. The second cube has volume 4³ = 64 cm³. The third cube has volume 5³ = 125 cm³. Adding these: 27 + 64 + 125 = 216 cm³. This is the volume of the new cube, so we need to find the edge length x where x³ = 216. Since 6³ = 216, the edge length is 6 cm. The total surface area of this new cube is 6 × 6² = 6 × 36 = 216 cm². At Rs.3.50 per cm², the cost of gold coating = 216 × 3.50 = Rs.756.
In simple words: Add up the volumes of the three small cubes. Find the cube root to get the edge of the new cube. Then use the surface area formula to find how much area needs coating, and multiply by the rate per square centimetre.

Exam Tip: Recognise that 27 + 64 + 125 = 216 = 6³ instantly — this is a key perfect-cube identity. Examiners often test whether you know that 1³ + 2³ + 3³ = 6³ and similar patterns.