Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 13 Pythagoras Theorem

Exercise 13(A)

Question 1. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Answer: Based on the Pythagorean principle, the square of the hypotenuse of a right-angled triangle equals the sum of the squares of the other two sides.
In this setup, let the ladder be represented by \( AB \), the ground by \( BC \), and the vertical wall by \( AC \).
Thus, in right-angled \( \Delta ACB \):
\( AB^2 = BC^2 + CA^2 \)
Substituting the values:
\( 13^2 = 5^2 + CA^2 \)
\( 169 = 25 + CA^2 \)
\( CA^2 = 169 - 25 \)
\( CA^2 = 144 \)
\( CA = \sqrt{144} = 12 \text{ m} \)
Consequently, the height reached on the wall by the ladder is 12 m.
In simple words: Use the Pythagorean formula with the ladder as the longest side. Subtract the square of the bottom distance from the square of the ladder's length to find the wall height.

Exam Tip: Always identify which side is the hypotenuse before using the theorem. The hypotenuse is always the longest side and lies directly opposite the right angle.

Question 2. A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.
Answer: To find the shortest distance from the start, we can model the journey using a right-angled triangle.
Let the starting location be \( A \). Moving Northwards by 40 m brings the person to point \( C \). Moving Westwards from \( C \) by 50 m brings him to \( B \).
Using the Pythagorean relation in right triangle \( \Delta BCA \):
\( AB^2 = BC^2 + CA^2 \)
\( AB^2 = 50^2 + 40^2 \)
\( AB^2 = 2500 + 1600 \)
\( AB^2 = 4100 \)
\( AB = \sqrt{4100} \approx 64.03 \text{ m} \)
Therefore, the man is approximately 64.03 m away from where he started.

In simple words: Imagine the path as two sides of a right triangle. The direct path back is the hypotenuse, which you calculate by squaring both distances, adding them together, and finding the square root.

Exam Tip: Clearly label your directions (North, South, East, West) to ensure you draw the right angle correctly. Here, the transition from North to West forms a 90-degree angle.

Question 3. In the figure: \( \angle PSQ = 90^\circ \), \( PQ = 10 \text{ cm} \), \( QS = 6 \text{ cm} \) and \( RQ = 9 \text{ cm} \). Calculate the length of PR.
Answer: First, let us find the length of the height \( PS \) by analyzing the smaller right-angled triangle \( \Delta PQS \) on the right.
Applying the Pythagoras theorem to \( \Delta PQS \):
\( PQ^2 = PS^2 + QS^2 \)
Substituting the known lengths:
\( 10^2 = PS^2 + 6^2 \)
\( 100 = PS^2 + 36 \)
\( PS^2 = 100 - 36 \)
\( PS^2 = 64 \)
\( PS = 8 \text{ cm} \)
Since points \( R, Q, \) and \( S \) lie in a straight line, the length of segment \( RS \) is the sum of \( RQ \) and \( QS \):
\( RS = RQ + QS = 9 \text{ cm} + 6 \text{ cm} = 15 \text{ cm} \)
Now we apply the theorem to the larger right triangle, \( \Delta PRS \):
\( PR^2 = RS^2 + PS^2 \)
\( PR^2 = 15^2 + 8^2 \)
\( PR^2 = 225 + 64 \)
\( PR^2 = 289 \)
\( PR = \sqrt{289} = 17 \text{ cm} \)
Thus, the length of segment \( PR \) is 17 cm.
In simple words: Use the smaller right triangle to calculate the height of the vertical line first. Then, add the two bottom horizontal parts together to get the total base of the large right triangle, and use both values to find the final long side.

Exam Tip: Be careful not to assume \( RQ \) is the base of a right-angled triangle. Only \( \Delta PQS \) and \( \Delta PRS \) have right angles, so calculations must be done using those triangles.

Question 4. The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and \( \angle ABD = \angle BCD = 90^\circ \). Calculate the length of AB.
Answer: We begin by focusing on the right-angled triangle \( \Delta BCD \), where the right angle is at \( C \).
By applying the Pythagorean theorem to this triangle:
\( DB^2 = DC^2 + BC^2 \)
\( DB^2 = 12^2 + 3^2 \)
\( DB^2 = 144 + 9 = 153 \)
Next, we look at the other right-angled triangle, \( \Delta ABD \), which has its right angle at \( B \).
Using the theorem for \( \Delta ABD \):
\( DA^2 = DB^2 + BA^2 \)
Substituting the values we know:
\( 13^2 = 153 + BA^2 \)
\( 169 = 153 + BA^2 \)
\( BA^2 = 169 - 153 \)
\( BA^2 = 16 \)
\( BA = \sqrt{16} = 4 \text{ cm} \)
Thus, the length of \( AB \) is 4 cm.
In simple words: First, use the bottom right triangle to find the square of the diagonal line. Then, use that diagonal value in the top right triangle to work backward and find the missing side.

Exam Tip: Keep \( DB^2 \) as 153 instead of calculating its decimal square root. Since you will square it again in the next step, using 153 directly prevents rounding errors.

Question 5. AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.
Answer: In an equilateral triangle, all three sides are equal. Thus, we have:
\( AB = AC = BC = 10 \text{ cm} \)
The perpendicular line \( AD \) from the top vertex to the base bisects the base \( BC \) into two equal segments:
\( BD = \frac{BC}{2} = \frac{10}{2} = 5 \text{ cm} \)
Now we analyze the right-angled triangle \( \Delta ABD \):
\( AB^2 = AD^2 + BD^2 \)
Substituting the values:
\( 10^2 = AD^2 + 5^2 \)
\( 100 = AD^2 + 25 \)
\( AD^2 = 100 - 25 \)
\( AD^2 = 75 \)
\( AD = \sqrt{75} \approx 8.66 \text{ cm} \)
Rounding to one decimal place, we get:
\( AD \approx 8.7 \text{ cm} \).
In simple words: Since the triangle is equilateral, the sides are all 10 cm. The line down the middle cuts the bottom side in half to make 5 cm, giving you a right triangle to calculate the height.

Exam Tip: Remember that in an equilateral triangle, the perpendicular bisector, altitude, and median from any vertex to the opposite side are all the same line.

Question 6. In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.
Answer: Let the length of the extended segment \( OC \) be \( x \text{ cm} \).
Since \( AO \perp BO \) (where \( O \) is on the line \( BC \) produced), both \( \Delta ABO \) and \( \Delta ACO \) are right-angled triangles at vertex \( O \).
For right triangle \( \Delta ACO \):
\( AC^2 = AO^2 + OC^2 \)
\( 3^2 = AO^2 + x^2 \)
\( AO^2 = 9 - x^2 \quad \text{--- (i)} \)
For right triangle \( \Delta ABO \):
Since \( OB = BC + OC = 6 + x \), we have:
\( AB^2 = AO^2 + OB^2 \)
\( 8^2 = AO^2 + (6 + x)^2 \)
\( AO^2 = 64 - (6 + x)^2 \quad \text{--- (ii)} \)
Equating the two expressions for \( AO^2 \) from (i) and (ii):
\( 9 - x^2 = 64 - (36 + 12x + x^2) \)
\( 9 - x^2 = 64 - 36 - 12x - x^2 \)
\( 9 = 28 - 12x \)
\( 12x = 28 - 9 \)
\( 12x = 19 \)
\( x = \frac{19}{12} = 1\frac{7}{12} \text{ cm} \)
Hence, the length of \( OC \) is \( 1\frac{7}{12} \text{ cm} \).
In simple words: Call the unknown extra base \( x \). Write equations for the vertical height from both the small and large right triangles, then set them equal to solve for \( x \).

Exam Tip: When a perpendicular is drawn outside the triangle, use the entire base length (original base plus the extension) for the larger right-angled triangle.

Question 7. In triangle ABC, AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm\(^2\). Find x.
Answer: Since triangle \( ABC \) is isosceles with \( AB = AC \), the perpendicular height \( AD \) dropped onto base \( BC \) splits \( BC \) into two equal halves:
\( BD = DC = \frac{BC}{2} = \frac{10}{2} = 5 \text{ cm} \)
Applying the Pythagorean theorem in right triangle \( \Delta ABD \):
\( AB^2 = AD^2 + BD^2 \)
\( x^2 = AD^2 + 5^2 \)
\( AD^2 = x^2 - 25 \)
\( AD = \sqrt{x^2 - 25} \)
The area of the triangle is given as 60 cm\(^2\):
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( 60 = \frac{1}{2} \times 10 \times \sqrt{x^2 - 25} \)
\( 60 = 5 \times \sqrt{x^2 - 25} \)
\( 12 = \sqrt{x^2 - 25} \)
Squaring both sides:
\( 144 = x^2 - 25 \)
\( x^2 = 144 + 25 \)
\( x^2 = 169 \)
\( x = \sqrt{169} = 13 \text{ cm} \)
Thus, the value of \( x \) is 13 cm.

In simple words: Find the height of the triangle using the area formula first. Then, use that height and half of the base in the Pythagoras formula to find the missing side \( x \).

Exam Tip: Using the property of an isosceles triangle to bisect the base is crucial. Always make sure to state this geometric property clearly to earn full method marks.

Question 8. If the sides of a triangle are in the ratio \( 1 : \sqrt{2} : 1 \), show that it is a right-angled triangle.
Answer: Let us define the lengths of the three sides of the triangle as \( x \), \( \sqrt{2}x \), and \( x \), where \( x \) is a positive multiplier.
Now, we calculate the squares of each of these sides:
First side squared: \( x^2 \)
Second side squared: \( x^2 \)
Hypotenuse (longest side) squared: \( (\sqrt{2}x)^2 = 2x^2 \)
Adding the squares of the two shorter sides:
\( x^2 + x^2 = 2x^2 \)
Since the sum of the squares of the two shorter sides equals the square of the longest side:
\( x^2 + x^2 = (\sqrt{2}x)^2 \)
This satisfies the converse of the Pythagorean theorem. Consequently, the given triangle is right-angled.
In simple words: Square all three sides. If the sum of the two smaller squares equals the largest square, then the triangle has a right angle.

Exam Tip: To show a triangle is right-angled, use the converse of Pythagoras theorem. Always identify the longest side first as the candidate for the hypotenuse.

Question 9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.
Answer: Let the two vertical poles be represented by \( AB = 6 \text{ m} \) and \( CD = 11 \text{ m} \). The distance between their bases is \( BC = 12 \text{ m} \).
Draw a horizontal construction line \( AE \) from the top of the shorter pole \( A \), meeting the taller pole \( CD \) perpendicularly at point \( E \).
This construction creates a rectangle \( ABCE \). Since opposite sides of a rectangle are equal:
\( AE = BC = 12 \text{ m} \)
\( EC = AB = 6 \text{ m} \)
Now, find the length of the remaining top section of the taller pole \( ED \):
\( ED = CD - EC = 11 \text{ m} - 6 \text{ m} = 5 \text{ m} \)
In the right-angled triangle \( \Delta AED \), applying Pythagoras theorem:
\( AD^2 = AE^2 + ED^2 \)
\( AD^2 = 12^2 + 5^2 \)
\( AD^2 = 144 + 25 \)
\( AD^2 = 169 \)
\( AD = \sqrt{169} = 13 \text{ m} \)
Therefore, the straight-line distance between the tops of the poles is 13 m.

In simple words: Draw a horizontal line from the top of the short pole to the tall pole to make a rectangle. This leaves a small right triangle at the top, which you can solve using Pythagoras.

Exam Tip: Drawing a correct auxiliary line is often the key to solving height and distance geometry questions. Clearly mention the creation of the rectangle and its equal sides.

Question 10. In the given figure, AB // CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.
Answer: Let us construct a perpendicular \( BM \) from vertex \( B \) to side \( CD \), meeting it at point \( M \).
This forms a rectangle \( ABMD \) because \( AB \parallel CD \) and \( AD \perp CD \).
Therefore:
\( DM = AB = 7 \text{ cm} \)
\( BM = AD \)
Since the total length of \( CD \) is 17 cm, we can find the remaining length \( MC \):
\( MC = CD - DM = 17 \text{ cm} - 7 \text{ cm} = 10 \text{ cm} \)
Now, let us find the height \( AD \) using the right-angled triangle \( \Delta BAD \) (with \( \angle A = 90^\circ \)):
\( BD^2 = AD^2 + AB^2 \)
\( 25^2 = AD^2 + 7^2 \)
\( 625 = AD^2 + 49 \)
\( AD^2 = 625 - 49 \)
\( AD^2 = 576 \)
\( AD = \sqrt{576} = 24 \text{ cm} \)
Since \( BM = AD \), we also have \( BM = 24 \text{ cm} \).
Finally, we apply Pythagoras theorem in the right triangle \( \Delta CMB \):
\( BC^2 = BM^2 + MC^2 \)
\( BC^2 = 24^2 + 10^2 \)
\( BC^2 = 576 + 100 \)
\( BC^2 = 676 \)
\( BC = \sqrt{676} = 26 \text{ cm} \)
Hence, the length of side \( BC \) is 26 cm.

In simple words: Break the shape into a rectangle on the left and a right triangle on the right. Find the vertical height from the left part, and then use that height with the right triangle's base to find the diagonal side \( BC \).

Exam Tip: Splitting a trapezoid into a rectangle and a right triangle is a very common technique in geometry problems. Clearly state which sides are equal due to the rectangle's properties.

Question 11. In the given figure, \( \angle B = 90^\circ \), \( XY \parallel BC \), \( AB = 12 \text{ cm} \), \( AY = 8 \text{ cm} \) and \( AX : XB = 1 : 2 \). Find the lengths of AC and BC.
Answer: We are given that \( AX : XB = 1 : 2 \). Let the common factor be \( n \).
So:
\( AX = n \text{ and } XB = 2n \)
Since \( AB = 12 \text{ cm} \):
\( AX + XB = 12 \)
\( n + 2n = 12 \)
\( 3n = 12 \)
\( \implies n = 4 \)
Therefore:
\( AX = 4 \text{ cm} \)
\( XB = 8 \text{ cm} \)
Since \( XY \parallel BC \), the triangles \( \Delta AXY \) and \( \Delta ABC \) are similar.
By the Thales's proportionality theorem:
\( \frac{AB}{AX} = \frac{AC}{AY} \)
\( \frac{12}{4} = \frac{AC}{8} \)
\( 3 = \frac{AC}{8} \)
\( AC = 24 \text{ cm} \)
Now, we use the Pythagorean theorem on right-angled triangle \( \Delta ABC \) (with \( \angle B = 90^\circ \)):
\( AC^2 = AB^2 + BC^2 \)
\( 24^2 = 12^2 + BC^2 \)
\( 576 = 144 + BC^2 \)
\( BC^2 = 576 - 144 \)
\( BC^2 = 432 \)
\( BC = \sqrt{432} = \sqrt{144 \times 3} = 12\sqrt{3} \text{ cm} \).
In simple words: Find the exact segments of side AB using the given ratio. Then use triangle similarity to find the length of AC. Finally, apply Pythagoras to calculate BC.

Exam Tip: When a line is parallel to one side of a triangle, it creates similar triangles. Remember to use the ratio of the full side (e.g., \( AB \)) to the upper segment (e.g., \( AX \)) to solve for the other side.

Question 12. In \( \Delta ABC \), \( \angle B = 90^\circ \). Find the sides of the triangle, if :
(i) AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm.
(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm.
Answer:

(i) In right-angled triangle \( \Delta ABC \) (with \( \angle B = 90^\circ \)), applying Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (x + 6)^2 = (x - 3)^2 + (x + 4)^2 \)
Expanding both sides:
\( x^2 + 12x + 36 = x^2 - 6x + 9 + x^2 + 8x + 16 \)
\( x^2 + 12x + 36 = 2x^2 + 2x + 25 \)
Rearranging into a standard quadratic form:
\( 2x^2 - x^2 + 2x - 12x + 25 - 36 = 0 \)
\( x^2 - 10x - 11 = 0 \)
Factoring the equation:
\( (x - 11)(x + 1) = 0 \)
Thus, \( x = 11 \) or \( x = -1 \).
Since a geometric side length cannot be negative, \( x = 11 \).
Calculating the sides:
\( AB = x - 3 = 11 - 3 = 8 \text{ cm} \)
\( BC = x + 4 = 11 + 4 = 15 \text{ cm} \)
\( AC = x + 6 = 11 + 6 = 17 \text{ cm} \)
(ii) In right-angled triangle \( \Delta ABC \) (with \( \angle B = 90^\circ \)), applying Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (4x + 5)^2 = x^2 + (4x + 4)^2 \)
Expanding the terms:
\( 16x^2 + 40x + 25 = x^2 + 16x^2 + 32x + 16 \)
Simplifying:
\( 16x^2 + 40x + 25 = 17x^2 + 32x + 16 \)
Rearranging into quadratic form:
\( x^2 - 8x - 9 = 0 \)
Factoring the quadratic equation:
\( (x - 9)(x + 1) = 0 \)
Thus, \( x = 9 \) or \( x = -1 \).
Since side lengths must be positive, \( x = 9 \).
Finding the lengths of the sides:
\( AB = x = 9 \text{ cm} \)
\( BC = 4x + 4 = 4(9) + 4 = 40 \text{ cm} \)
\( AC = 4x + 5 = 4(9) + 5 = 41 \text{ cm} \)
In simple words: Put the given expressions into the Pythagoras formula. Solve the resulting quadratic equation, pick the positive answer, and plug it back to get the final side lengths.

Exam Tip: Quadratic equations in geometry always produce two roots. You must explicitly discard the negative root because lengths are strictly positive.

Exercise 13(B)

Question 1. In the figure, given below, \( AD \perp BC \). Prove that: \( c^2 = a^2 + b^2 - 2ax \).
Answer: Let us examine the two right-angled triangles formed by the altitude \( AD \).
In the right-angled triangle \( \Delta ACD \), applying the Pythagoras theorem:
\( AC^2 = AD^2 + CD^2 \)
\( b^2 = h^2 + x^2 \)
Expressing \( h^2 \) in terms of other variables:
\( h^2 = b^2 - x^2 \quad \text{--- (i)} \)
In the right-angled triangle \( \Delta ABD \), we also apply the theorem:
\( AB^2 = AD^2 + BD^2 \)
Here, the length of \( BD \) is \( a - x \):
\( c^2 = h^2 + (a - x)^2 \)
Expressing \( h^2 \) in terms of other variables:
\( h^2 = c^2 - (a - x)^2 \quad \text{--- (ii)} \)
Equating the two expressions for \( h^2 \) from (i) and (ii):
\( b^2 - x^2 = c^2 - (a - x)^2 \)
Expanding the bracket:
\( b^2 - x^2 = c^2 - (a^2 - 2ax + x^2) \)
\( b^2 - x^2 = c^2 - a^2 + 2ax - x^2 \)
Cancel \( -x^2 \) from both sides of the equation:
\( b^2 = c^2 - a^2 + 2ax \)
Rearranging the terms to solve for \( c^2 \):
\( c^2 = a^2 + b^2 - 2ax \)
Hence Proved.
In simple words: Write equations for the vertical height of both right triangles, set them equal to each other, expand the brackets, and rearrange the letters to prove the statement.

Exam Tip: To prove algebraic relations in triangles with altitudes, express the shared perpendicular altitude \( h^2 \) in two different ways using Pythagoras theorem, and then equate them.

Question 2. In equilateral triangle ABC, \( AD \perp BC \) and BC = x cm. Find, in terms of x, the length of AD.
Answer: In an equilateral triangle \( ABC \), all sides have the same length:
\( AB = AC = BC = x \text{ cm} \)
The perpendicular height \( AD \) divides the base \( BC \) into two equal segments:
\( BD = DC = \frac{BC}{2} = \frac{x}{2} \text{ cm} \)
Now we apply Pythagoras theorem in the right triangle \( \Delta ADC \):
\( AC^2 = AD^2 + DC^2 \)
\( x^2 = AD^2 + \left(\frac{x}{2}\right)^2 \)
\( x^2 = AD^2 + \frac{x^2}{4} \)
\( AD^2 = x^2 - \frac{x^2}{4} \)
\( AD^2 = \frac{4x^2 - x^2}{4} \)
\( AD^2 = \frac{3x^2}{4} \)
Taking the positive square root:
\( AD = \frac{\sqrt{3}}{2} x \text{ cm} \)
Thus, the length of \( AD \) is \( \frac{\sqrt{3}}{2} x \text{ cm} \).

In simple words: Since the triangle is equilateral, the sides are all \( x \). The vertical line cuts the bottom in half to make \( \frac{x}{2} \). Use Pythagoras to find the height in terms of \( x \).

Exam Tip: In any equilateral triangle of side \( s \), the altitude is always \( \frac{\sqrt{3}}{2}s \). This is an extremely useful direct formula to remember for competitive exams.

Question 3. In a right-angled triangle ABC, right-angled at B. M is a point on BC. Prove that: \( AM^2 + BC^2 = AC^2 + BM^2 \).
Answer: In the right-angled triangle \( \Delta ABC \), the angle at \( B \) is \( 90^\circ \). Point \( M \) is located on segment \( BC \).
Let us apply Pythagoras theorem to the smaller right-angled triangle \( \Delta ABM \):
\( AM^2 = AB^2 + BM^2 \)
Expressing the common side \( AB^2 \):
\( AB^2 = AM^2 - BM^2 \quad \text{--- (i)} \)
Next, we apply Pythagoras theorem to the main right-angled triangle \( \Delta ABC \):
\( AC^2 = AB^2 + BC^2 \)
Expressing \( AB^2 \):
\( AB^2 = AC^2 - BC^2 \quad \text{--- (ii)} \)
By equating both expressions for \( AB^2 \) from (i) and (ii):
\( AM^2 - BM^2 = AC^2 - BC^2 \)
Rearranging the terms:
\( AM^2 + BC^2 = AC^2 + BM^2 \)
Hence Proved.
In simple words: Write Pythagoras equations for both right triangles sharing the vertical side AB, rearrange them to isolate AB on one side, then set them equal to each other to complete the proof.

Exam Tip: When proving equations where terms are added on both sides, look for a shared side in the triangles. Expressing that shared side in two ways and equating them is the standard approach.

Question 4. M and N are the mid-points of the sides QR and PQ respectively of a triangle PQR, right-angled at Q. Prove that:
(i) \( PM^2 + RN^2 = 5 MN^2 \)
(ii) \( 4PM^2 = 4PQ^2 + QR^2 \)
(iii) \( 4RN^2 = 4RQ^2 + PQ^2 \)
(iv) \( 4(PM^2 + RN^2) = 5PR^2 \)
Answer: Since \( M \) is the mid-point of \( QR \), and \( N \) is the mid-point of \( PQ \), we have:
\( PQ = 2NQ \text{ and } QR = 2QM \)
(i) In right-angled triangle \( \Delta PQM \):
\( PM^2 = PQ^2 + QM^2 \)
\( PM^2 = (2NQ)^2 + QM^2 \)
\( PM^2 = 4NQ^2 + QM^2 \quad \text{--- (a)} \)
In right-angled triangle \( \Delta RNQ \):
\( RN^2 = NQ^2 + QR^2 \)
\( RN^2 = NQ^2 + (2QM)^2 \)
\( RN^2 = NQ^2 + 4QM^2 \quad \text{--- (b)} \)
Adding equations (a) and (b):
\( PM^2 + RN^2 = (4NQ^2 + QM^2) + (NQ^2 + 4QM^2) \)
\( PM^2 + RN^2 = 5NQ^2 + 5QM^2 \)
\( PM^2 + RN^2 = 5(NQ^2 + QM^2) \)
In the right-angled triangle \( \Delta NMQ \):
\( MN^2 = NQ^2 + QM^2 \)
Substituting this into our equation:
\( PM^2 + RN^2 = 5MN^2 \)
Hence Proved.
(ii) In right-angled triangle \( \Delta PQM \):
\( PM^2 = PQ^2 + QM^2 \)
Multiplying both sides by 4:
\( 4PM^2 = 4PQ^2 + 4QM^2 \)
Since \( 4QM^2 = (2QM)^2 = QR^2 \):
\( 4PM^2 = 4PQ^2 + QR^2 \)
Hence Proved.
(iii) In right-angled triangle \( \Delta RNQ \):
\( RN^2 = NQ^2 + QR^2 \)
Multiplying both sides by 4:
\( 4RN^2 = 4NQ^2 + 4QR^2 \)
Since \( 4NQ^2 = (2NQ)^2 = PQ^2 \):
\( 4RN^2 = PQ^2 + 4QR^2 \)
Hence Proved.
(iv) Adding the results from (ii) and (iii):
\( 4PM^2 + 4RN^2 = (4PQ^2 + QR^2) + (PQ^2 + 4QR^2) \)
\( 4(PM^2 + RN^2) = 5PQ^2 + 5QR^2 \)
\( 4(PM^2 + RN^2) = 5(PQ^2 + QR^2) \)
In right-angled triangle \( \Delta PQR \):
\( PR^2 = PQ^2 + QR^2 \)
Substituting this back into our equation:
\( 4(PM^2 + RN^2) = 5PR^2 \)
Hence Proved.
In simple words: Use the fact that M and N split the sides of the right-angled triangle in half. Apply Pythagoras to the smaller triangles inside the shape and combine the equations to prove each of the four statements.

Exam Tip: For multi-part proofs involving mid-points in a right triangle, always write out the basic Pythagoras relations for all possible right triangles first (\(\Delta PQM\), \(\Delta RNQ\), \(\Delta MNQ\), and \(\Delta PQR\)). This makes substitution straightforward.

Exercise 13(B)

Question 5. In triangle ABC, \(\angle B = 90^\circ\) and D is the mid-point of BC. Prove that: \(AC^2 = AD^2 + 3CD^2\).
Answer:
By the Pythagorean theorem, in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Since \(D\) is the midpoint of \(BC\), we have:
\(BD = CD = \frac{1}{2}BC\) (which means \(BC = 2CD\))
First, in the right-angled triangle \(\triangle ABD\):
\(AD^2 = AB^2 + BD^2\)
\( \implies AB^2 = AD^2 - BD^2 \quad \text{--- (1)} \)
Next, in the larger right-angled triangle \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2\)
\( \implies AB^2 = AC^2 - BC^2 \quad \text{--- (2)} \)
By equating the expressions for \(AB^2\) from (1) and (2), we get:
\(AC^2 - BC^2 = AD^2 - BD^2\)
\( \implies AC^2 = AD^2 - BD^2 + BC^2 \)
Now, substitute \(BD = CD\) and \(BC = 2CD\) into the equation:
\(AC^2 = AD^2 - CD^2 + (2CD)^2\)
\( \implies AC^2 = AD^2 - CD^2 + 4CD^2 \)
\( \implies AC^2 = AD^2 + 3CD^2 \)
Hence proved.
In simple words: We apply the Pythagorean theorem to two different right-angled triangles in the diagram, \(ABD\) and \(ABC\). By writing the common side \(AB^2\) in two different ways, we can equate them and solve to get the required relation.

Exam Tip: Remember to write the substitution \(BC = 2CD\) clearly, as examiners look for this key logical step when grading.

Question 6. In a rectangle ABCD, prove that: \(AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2\).
Answer:
By definition, all four angles of a rectangle \(ABCD\) are right angles (\(90^\circ\)).
First, let us consider the right-angled triangle \(\triangle ACD\). Applying the Pythagorean theorem:
\(AC^2 = DA^2 + CD^2 \quad \text{--- (1)} \)
Similarly, for the right-angled triangle \(\triangle BDC\), applying the Pythagorean theorem:
\(BD^2 = BC^2 + CD^2\)
Since the opposite sides of a rectangle are equal, we can replace \(CD\) with \(AB\) (\(CD = AB\)):
\( \implies BD^2 = BC^2 + AB^2 \quad \text{--- (2)} \)
Adding the equations (1) and (2) together, we obtain:
\(AC^2 + BD^2 = (DA^2 + CD^2) + (BC^2 + AB^2)\)
\( \implies AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2 \)
Hence proved.
In simple words: We apply the Pythagorean theorem to two right-angled triangles created by the diagonals. Summing these two equations and using the fact that opposite sides of a rectangle are equal gives the desired proof.

Exam Tip: Be sure to state that \(\angle C = 90^\circ\) and \(\angle D = 90^\circ\) to formally justify using the Pythagorean theorem.

Question 7. In quadrilateral ABCD, \(\angle B = 90^\circ\) and \(\angle D = 90^\circ\). Prove that: \(2AC^2 - AB^2 = AD^2 + DC^2 + BC^2\).
Answer:
In the given quadrilateral \(ABCD\), we have \(\angle B = 90^\circ\) and \(\angle D = 90^\circ\). Thus, both \(\triangle ABC\) and \(\triangle ADC\) are right-angled triangles.
Applying the Pythagorean theorem in \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2\)
\( \implies AB^2 = AC^2 - BC^2 \quad \text{--- (1)} \)
Applying the Pythagorean theorem in \(\triangle ADC\):
\(AC^2 = AD^2 + DC^2 \quad \text{--- (2)} \)
Now, let us simplify the Left-Hand Side (LHS) of the relation to be proved:
\(\text{LHS} = 2AC^2 - AB^2\)
Substitute the value of \(AB^2\) from equation (1):
\(\text{LHS} = 2AC^2 - (AC^2 - BC^2)\)
\( \implies \text{LHS} = 2AC^2 - AC^2 + BC^2 \)
\( \implies \text{LHS} = AC^2 + BC^2 \)
Now, substitute the value of \(AC^2\) from equation (2):
\(\text{LHS} = (AD^2 + DC^2) + BC^2\)
\( \implies \text{LHS} = AD^2 + DC^2 + BC^2 = \text{RHS} \)
Hence proved.
In simple words: Since the diagonal \(AC\) is the hypotenuse for both right-angled triangles, we can write equations for both. Substituting these equations into the left-hand side of our expression simplifies it directly to the right-hand side.

Exam Tip: Highlighting the common hypotenuse \(AC\) makes the proof straightforward and shows the examiner you understand the geometric connection.

Question 8. O is any point inside a rectangle ABCD. Prove that: \(OB^2 + OD^2 = OC^2 + OA^2\).
Answer:
Let \(ABCD\) be a rectangle and \(O\) be any arbitrary point inside it.
Construct perpendiculars from \(O\) to the four sides of the rectangle:
\(OE \perp AB\) (with \(E\) on \(AB\)), \(OF \perp BC\) (with \(F\) on \(BC\)), \(OG \perp CD\) (with \(G\) on \(CD\)), and \(OH \perp DA\) (with \(H\) on \(DA\)).
This construction forms four smaller rectangles: \(AEOH\), \(EBFO\), \(OFCG\), and \(OHDG\).
Since opposite sides of a rectangle are equal, we have:
\(OH = AE\), \(CG = EB\), \(EO = AH\), and \(OG = HD\).
Now, we apply the Pythagorean theorem to the relevant right-angled triangles:
1. In right-angled triangle \(\triangle AHO\):
\(OA^2 = AH^2 + OH^2\)
\( \implies OA^2 = AH^2 + AE^2 \quad \text{--- (1)} \) (since \(OH = AE\))
2. In right-angled triangle \(\triangle CGO\):
\(OC^2 = CG^2 + OG^2\)
\( \implies OC^2 = EB^2 + HD^2 \quad \text{--- (2)} \) (since \(CG = EB\) and \(OG = HD\))
3. In right-angled triangle \(\triangle BEO\):
\(OB^2 = EO^2 + BE^2\)
\( \implies OB^2 = AH^2 + BE^2 \quad \text{--- (3)} \) (since \(EO = AH\))
4. In right-angled triangle \(\triangle DHO\):
\(OD^2 = HD^2 + OH^2\)
\( \implies OD^2 = HD^2 + AE^2 \quad \text{--- (4)} \) (since \(OH = AE\))
Adding equations (1) and (2) together:
\(OA^2 + OC^2 = AH^2 + AE^2 + EB^2 + HD^2 \quad \text{--- (5)} \)
Adding equations (3) and (4) together:
\(OB^2 + OD^2 = AH^2 + BE^2 + HD^2 + AE^2 \quad \text{--- (6)} \)
Comparing equations (5) and (6), we find that their right-hand sides are identical:
\( \implies OA^2 + OC^2 = OB^2 + OD^2 \)
Hence proved.

In simple words: By drawing perpendicular lines from \(O\) to each of the sides, we divide the rectangle into smaller parts. We then use the Pythagorean theorem on the four triangles meeting at \(O\) and match equal sides of the small rectangles to prove the formula.

Exam Tip: Don't forget to explicitly write down the equal segment pairs like \(OH = AE\) and \(CG = EB\) before using them in your equations, so your proof is fully justified.

Question 9. O is any point in the interior of a triangle ABC. Perpendiculars OP, OQ, and OR are drawn to the sides BC, CA, and AB respectively. Prove that: \(AR^2 + BP^2 + CQ^2 = AQ^2 + CP^2 + BR^2\).
Answer:
To solve this, let us connect the interior point \(O\) to the vertices \(A\), \(B\), and \(C\).
Since \(OP \perp BC\), \(OQ \perp CA\), and \(OR \perp AB\), several right-angled triangles are formed around point \(O\).
Applying the Pythagorean theorem to these triangles:
In right-angled triangle \(\triangle ARO\):
\(AO^2 = AR^2 + OR^2\)
\( \implies AR^2 = AO^2 - OR^2 \quad \text{--- (1)} \)
Similarly, we write the relations for the remaining right-angled triangles:
In right-angled triangle \(\triangle BPO\):
\(BP^2 = BO^2 - OP^2 \quad \text{--- (2)} \)
In right-angled triangle \(\triangle COQ\):
\(CQ^2 = CO^2 - OQ^2 \quad \text{--- (3)} \)
In right-angled triangle \(\triangle AOQ\):
\(AQ^2 = AO^2 - OQ^2 \quad \text{--- (4)} \)
In right-angled triangle \(\triangle CPO\):
\(CP^2 = CO^2 - OP^2 \quad \text{--- (5)} \)
In right-angled triangle \(\triangle BRO\):
\(BR^2 = BO^2 - OR^2 \quad \text{--- (6)} \)
Adding equations (1), (2), and (3):
\(AR^2 + BP^2 + CQ^2 = (AO^2 - OR^2) + (BO^2 - OP^2) + (CO^2 - OQ^2)\)
\( \implies AR^2 + BP^2 + CQ^2 = AO^2 + BO^2 + CO^2 - OP^2 - OQ^2 - OR^2 \quad \text{--- (7)} \)
Adding equations (4), (5), and (6):
\(AQ^2 + CP^2 + BR^2 = (AO^2 - OQ^2) + (CO^2 - OP^2) + (BO^2 - OR^2)\)
\( \implies AQ^2 + CP^2 + BR^2 = AO^2 + BO^2 + CO^2 - OP^2 - OQ^2 - OR^2 \quad \text{--- (8)} \)
Comparing (7) and (8), we find their right-hand sides are equal:
\( \implies AR^2 + BP^2 + CQ^2 = AQ^2 + CP^2 + BR^2 \)
Hence proved.
In simple words: Connecting point \(O\) to the corners creates six small right-angled triangles. By writing the square of each side of the proof using the Pythagorean theorem, we show that both sides of the proof add up to the exact same total.

Exam Tip: Be sure to write down the construction of auxiliary lines \(OA\), \(OB\), and \(OC\) in your exam paper, as they form the right-angled triangles necessary for the proof.

Question 10. Diagonals of rhombus ABCD intersect each other at point O. Prove that: \(OA^2 + OC^2 = 2AD^2 - \frac{BD^2}{2}\).
Answer:
In a rhombus \(ABCD\), the diagonals are perpendicular to each other at their point of intersection \(O\).
Therefore, we have \(\angle AOD = \angle COD = 90^\circ\), which means \(\triangle AOD\) and \(\triangle COD\) are right-angled triangles.
Applying the Pythagorean theorem in \(\triangle AOD\):
\(AD^2 = OA^2 + OD^2\)
\( \implies OA^2 = AD^2 - OD^2 \quad \text{--- (1)} \)
Applying the Pythagorean theorem in \(\triangle COD\):
\(CD^2 = OC^2 + OD^2\)
\( \implies OC^2 = CD^2 - OD^2 \quad \text{--- (2)} \)
Now, we add the equations (1) and (2) to get the Left-Hand Side (LHS) of our expression:
\(\text{LHS} = OA^2 + OC^2\)
\(\text{LHS} = (AD^2 - OD^2) + (CD^2 - OD^2)\)
\( \implies \text{LHS} = AD^2 + CD^2 - 2OD^2 \)
Since \(ABCD\) is a rhombus, all its sides are equal (\(CD = AD\)), and its diagonals bisect each other (\(OD = \frac{BD}{2}\)). Substituting these relations:
\(\text{LHS} = AD^2 + AD^2 - 2\left(\frac{BD}{2}\right)^2\)
\( \implies \text{LHS} = 2AD^2 - 2\left(\frac{BD^2}{4}\right) \)
\( \implies \text{LHS} = 2AD^2 - \frac{BD^2}{2} = \text{RHS} \)
Hence proved.

In simple words: The diagonals of a rhombus always meet at right angles. By using the Pythagorean theorem on two right triangles that share a side, we can combine their formulas and substitute the properties of a rhombus to get the result.

Exam Tip: Remember to explicitly state that all sides of a rhombus are equal (\(AB = BC = CD = DA\)) and that diagonals bisect each other to justify your algebraic substitutions.

Question 11. In the figure, AB = BC and AD is perpendicular to CD. Prove that: \(AC^2 = 2BC \cdot DC\).
Answer:
Given that \(AB = BC\) and \(AD \perp CD\), we have two right-angled triangles, \(\triangle ADB\) and \(\triangle ADC\), with the right angle at \(D\).
First, applying the Pythagorean theorem in \(\triangle ADB\):
\(AB^2 = AD^2 + BD^2\)
\( \implies AD^2 = AB^2 - BD^2 \quad \text{--- (1)} \)
Next, applying the Pythagorean theorem in \(\triangle ADC\):
\(AC^2 = AD^2 + DC^2 \quad \text{--- (2)} \)
Since the point \(B\) lies on the line segment \(DC\), we can write:
\(DC = DB + BC\)
Now, substitute the value of \(AD^2\) from equation (1) and replace \(DC\) with \((DB + BC)\) in equation (2):
\(AC^2 = (AB^2 - BD^2) + (DB + BC)^2\)
Expanding the quadratic term:
\(AC^2 = AB^2 - BD^2 + DB^2 + BC^2 + 2 \cdot DB \cdot BC\)
\( \implies AC^2 = AB^2 + BC^2 + 2 \cdot DB \cdot BC \)
Since it is given that \(AB = BC\), substitute \(BC\) in place of \(AB\):
\(AC^2 = BC^2 + BC^2 + 2 \cdot DB \cdot BC\)
\( \implies AC^2 = 2BC^2 + 2 \cdot DB \cdot BC \)
Factor out \(2BC\) from the expression on the right-hand side:
\(AC^2 = 2BC(BC + DB)\)
Substitute \(BC + DB = DC\) back into the equation:
\( \implies AC^2 = 2BC \cdot DC \)
Hence proved.
In simple words: We apply the Pythagorean theorem to triangles \(\triangle ADB\) and \(\triangle ADC\). Since point \(B\) divides the base, we can write the base as a sum of two parts, expand it, and use \(AB = BC\) to simplify the terms to the final result.

Exam Tip: Be very careful when expanding \((DB + BC)^2\); make sure to include the \(2 \cdot DB \cdot BC\) term, which is essential to reaching the final proven equation.

Question 12. In an isosceles triangle ABC; AB = AC and D is a point on BC produced. Prove that: \(AD^2 = AC^2 + BD \cdot CD\).
Answer:
Let \(ABC\) be an isosceles triangle with \(AB = AC\), and let \(D\) be a point on the line segment \(BC\) produced.
Construct the altitude \(AE \perp BC\).
First, let us apply the Pythagorean theorem in the right-angled triangle \(\triangle AED\):
\(AD^2 = AE^2 + ED^2\)
Since the point \(C\) lies on the segment \(ED\), we have \(ED = EC + CD\). Thus:
\(AD^2 = AE^2 + (EC + CD)^2 \quad \text{--- (1)} \)
Next, applying the Pythagorean theorem in the right-angled triangle \(\triangle AEC\):
\(AC^2 = AE^2 + EC^2\)
\( \implies AE^2 = AC^2 - EC^2 \quad \text{--- (2)} \)
Substitute the expression for \(AE^2\) from equation (2) into equation (1):
\(AD^2 = (AC^2 - EC^2) + (EC + CD)^2\)
Expanding the squared term:
\(AD^2 = AC^2 - EC^2 + EC^2 + CD^2 + 2 \cdot EC \cdot CD\)
\( \implies AD^2 = AC^2 + CD^2 + 2 \cdot EC \cdot CD \)
Factor out \(CD\) from the last two terms:
\(AD^2 = AC^2 + CD(CD + 2EC) \quad \text{--- (3)} \)
In an isosceles triangle, the perpendicular drawn from the vertex to the base bisects the base. Therefore, \(E\) is the midpoint of \(BC\), meaning:
\(BC = 2EC\)
From the diagram, the total length \(BD\) is:
\(BD = BC + CD\)
Substitute \(BC = 2EC\) into this relation:
\(BD = 2EC + CD\)
Now, substitute \(2EC + CD = BD\) into equation (3):
\( \implies AD^2 = AC^2 + BD \cdot CD \)
Hence proved.

In simple words: We draw a line perpendicular to the base, creating right-angled triangles. By applying the Pythagorean theorem and using the property that the altitude bisects the base of an isosceles triangle, we can simplify the equation to find the required relation.

Exam Tip: Be sure to state clearly that the altitude \(AE\) bisects the base \(BC\) in the isosceles triangle \(ABC\), which is the essential geometric property used for the final simplification step.

Question 13. In an isosceles triangle ABC; \(\angle A = 90^\circ\), CA = AB and D is a point on AB produced. Prove that: \(DC^2 - BD^2 = 2AB \cdot AD\).
Answer:
In the right-angled triangle \(\triangle ACD\), since \(\angle A = 90^\circ\), applying the Pythagorean theorem gives:
\(CD^2 = AC^2 + AD^2\)
Since \(ABC\) is an isosceles triangle with \(CA = AB\), we can substitute \(AB\) for \(AC\):
\( \implies CD^2 = AB^2 + AD^2 \quad \text{--- (1)} \)
Since \(D\) is a point on \(AB\) produced, the points \(A\), \(B\), and \(D\) are collinear, and we have:
\(AD = AB + BD\)
\( \implies BD = AD - AB \)
Squaring both sides of this equation, we get:
\(BD^2 = (AD - AB)^2\)
\( \implies BD^2 = AD^2 + AB^2 - 2AB \cdot AD \quad \text{--- (2)} \)
Now, subtract equation (2) from equation (1):
\(CD^2 - BD^2 = (AB^2 + AD^2) - (AD^2 + AB^2 - 2AB \cdot AD)\)
\( \implies CD^2 - BD^2 = AB^2 + AD^2 - AD^2 - AB^2 + 2AB \cdot AD \)
Simplifying the right-hand side:
\( \implies DC^2 - BD^2 = 2AB \cdot AD \)
Hence proved.
In simple words: We write the Pythagorean equation for the right triangle \(\triangle ACD\) and replace \(AC\) with \(AB\). Then, writing \(BD\) as \(AD - AB\) and squaring it allows us to subtract the equations and get the required formula directly.

Exam Tip: Substituting the equal side \(AC = AB\) immediately at the start of the proof makes the algebraic reduction much simpler and prevents calculation errors.

Question 14. In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that: \(BD^2 - CD^2 = 2CD \cdot AD\).
Answer:
In right-angled triangle \(\triangle ABD\), since \(BD \perp AC\), applying the Pythagorean theorem gives:
\(AB^2 = AD^2 + BD^2\)
\( \implies AD^2 = AB^2 - BD^2 \quad \text{--- (1)} \)
From the figure, the side \(AC\) is composed of two segments, \(AD\) and \(DC\):
\(AC = AD + CD\)
Squaring both sides of this relation:
\(AC^2 = (AD + CD)^2\)
\( \implies AC^2 = AD^2 + CD^2 + 2 \cdot AD \cdot CD \)
Substitute the value of \(AD^2\) from equation (1):
\(AC^2 = (AB^2 - BD^2) + CD^2 + 2 \cdot AD \cdot CD\)
Since it is given that \(AB = AC\), we substitute \(AC\) in place of \(AB\):
\(AC^2 = AC^2 - BD^2 + CD^2 + 2 \cdot AD \cdot CD\)
Subtract \(AC^2\) from both sides of the equation:
\(0 = -BD^2 + CD^2 + 2 \cdot CD \cdot AD\)
Rearranging the terms:
\( \implies BD^2 - CD^2 = 2CD \cdot AD \)
Hence proved.
In simple words: We write the total length \(AC\) as \(AD + CD\) and square it. By substituting the Pythagorean formula for triangle \(\triangle ABD\) and using the given fact that \(AB = AC\), we can cancel out terms on both sides to find the relation.

Exam Tip: Be sure to write the full binomial expansion of \((AD + CD)^2\) carefully, and clearly note where \(AB = AC\) is substituted to cancel the \(AC^2\) terms.

Question 15. In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3. Prove that: \(2AC^2 = 2AB^2 + BC^2\).
Answer:
Given that \(AD \perp BC\) and \(D\) divides \(BC\) in the ratio \(1:3\), we have:
\(BD : CD = 1 : 3\)
Since the total length is \(BC = BD + CD\), we can write \(BD\) and \(CD\) as fractions of \(BC\):
\(BD = \frac{1}{4}BC \quad \text{and} \quad CD = \frac{3}{4}BC\)
Now, apply the Pythagorean theorem in the right-angled triangles \(\triangle ACD\) and \(\triangle ABD\):
In \(\triangle ACD\):
\(AC^2 = AD^2 + CD^2 \quad \text{--- (1)} \)
In \(\triangle ABD\):
\(AB^2 = AD^2 + BD^2 \quad \text{--- (2)} \)
Subtract equation (2) from equation (1) to eliminate the common term \(AD^2\):
\(AC^2 - AB^2 = CD^2 - BD^2\)
Substitute the fractional values of \(BD\) and \(CD\) in terms of \(BC\):
\(AC^2 - AB^2 = \left(\frac{3}{4}BC\right)^2 - \left(\frac{1}{4}BC\right)^2\)
\( \implies AC^2 - AB^2 = \frac{9}{16}BC^2 - \frac{1}{16}BC^2 \)
\( \implies AC^2 - AB^2 = \frac{8}{16}BC^2 \)
\( \implies AC^2 - AB^2 = \frac{1}{2}BC^2 \)
Multiply both sides by 2:
\(2(AC^2 - AB^2) = BC^2\)
\( \implies 2AC^2 - 2AB^2 = BC^2 \)
Rearranging the terms:
\( \implies 2AC^2 = 2AB^2 + BC^2 \)
Hence proved.
In simple words: We express the segments \(BD\) and \(CD\) as fractions of the entire base \(BC\) according to the given ratio. By subtracting the Pythagorean equations for the two triangles, we eliminate \(AD\) and simplify the terms to obtain the required relation.

Exam Tip: Be careful when writing the ratio fractions. Since the ratio is \(1:3\), the total parts are \(1 + 3 = 4\), so the segments are \(\frac{1}{4}\) and \(\frac{3}{4}\) of \(BC\).