CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 02

Surface Areas and Volumes - Important Formulas

Additional Formula Relations:

• 9. Diagonal of cuboid = \( \sqrt{l^2 + b^2 + h^2} \)
• 10. Diagonal of Cube = \( \sqrt{3}l \)

Level-I

Question. In a right circular cone the cross section made by a plane parallel to the base is a.
(i) Circle
(ii) Frustum of a cone
(iii) Sphere
(iv) Semi sphere
Answer: (i) Circle
In simple words: Slicing a cone horizontally, completely parallel to its base, always creates a boundary that forms a perfect circular shape.

Exam Tip: Always visualize the cross-section parallel to the flat face. For any symmetric vertical shape with a circular base, a horizontal cut yields a circle.

Question. The radius and height of cylinder are in the ratio 5:7 and its volume is 550 cm3. Its radius is
(i) 1 cm
(ii) 7 cm
(iii) 5 cm
(iv) 6 cm
Answer: (iii) 5 cm
In simple words: Represent the radius as 5x and the height as 7x. Solving the volume equation gives x equal to 1, meaning the radius is 5 cm.

Exam Tip: Set up variables using a common ratio factor \( x \). Write down the equation \( V = \pi r^2 h \) clearly before plugging in numbers to prevent algebraic mistakes.

Question. A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes.
(i) 1:2:3
(ii) 3:1:3
(iii) 3:1:2
(iv) 2/3:1/3:1
Answer: (iii) 3:1:2
In simple words: For a hemisphere, the height is the same as its radius. Putting this radius-to-height relationship into the volume formulas yields a ratio of 3 to 1 to 2.

Exam Tip: Remember that a hemisphere's height must equal its radius. Using \( h = r \) across all formulas simplifies the math immediately.

Question. If surface areas of two spheres are in the ratio 4:9 then the ratio of their volumes is :
(i) 16/27
(ii) 4/27
(iii) 8/27
(iv) 9/27
Answer: (iii) 8/27
In simple words: Take the square root of the surface area ratio to get the radii ratio, which is 2 to 3. Cubing this gives the volume ratio of 8 to 27.

Exam Tip: The ratio of surface areas represents \( r_1^2:r_2^2 \), whereas the volume ratio represents \( r_1^3:r_2^3 \). Use this relationship directly to save calculation time.

Question. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube:
(i) 4:π
(ii) π:2
(iii) 3:π
(iv) 6:π
Answer: (iv) 6:π
In simple words: The sphere's diameter is equal to the edge length of the cube. Dividing the cube's volume by the sphere's volume leaves us with the ratio 6 to pi.

Exam Tip: A sphere inscribed inside a cube always has its diameter equal to the side length of the cube. Remember to divide the diameter by 2 to get the radius before using the volume formula.

Level-II

Question. The slant height of a frustum of a cone is 10 cm. If the height of the frustum is 8 cm, then find the difference of the radius of its two circular ends.
Answer: Let the radii of the two ends of the frustum be \( R \) and \( r \). We are given the slant height \( l = 10\text{ cm} \) and the height \( h = 8\text{ cm} \). We use the frustum relationship formula:
\[ l^2 = h^2 + (R - r)^2 \]
\[ 10^2 = 8^2 + (R - r)^2 \]
\[ 100 = 64 + (R - r)^2 \]
\[ (R - r)^2 = 36 \]
\[ R - r = \sqrt{36} = 6\text{ cm} \]
Therefore, the difference of the radii is \( 6\text{ cm} \).
In simple words: We find the missing side of the right triangle formed by the frustum's height and slant height, which shows that the difference between the top and bottom radii is 6 cm.

Exam Tip: Do not try to solve for individual radii \( R \) and \( r \). Treat \( (R - r) \) as a single variable and solve for it directly using the formula.

Question. A solid metallic sphere of radius 12 cm is melted and recast into a number of small cones each of radius 4 cm and height 3 cm . Find the number of cones so formed.
Answer: The total volume remains constant during melting. Let \( N \) be the number of small cones. The volume of the sphere is:
\[ V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (12)^3\text{ cm}^3 \]
The volume of each individual cone is:
\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4)^2 (3)\text{ cm}^3 \]
We calculate the number of cones:
\[ N = \frac{V_{\text{sphere}}}{V_{\text{cone}}} = \frac{\frac{4}{3}\pi \times 12 \times 12 \times 12}{\frac{1}{3}\pi \times 4 \times 4 \times 3} \]
\[ N = \frac{4 \times 12 \times 12 \times 12}{16 \times 3} = \frac{6912}{48} = 144 \]
So, \( 144 \) cones are created.
In simple words: Divide the total volume of the large sphere by the volume of a single small cone. The division gives a total of 144 cones.

Exam Tip: Never calculate the actual decimal value of \( \pi \). Keeping the calculations in terms of \( \pi \) allows it to cancel out cleanly during division.

Question. How many spherical lead shots of radius 2 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Answer: Let \( N \) be the total number of spherical lead shots. The volume of the lead cube is:
\[ V_{\text{cube}} = a^3 = 44 \times 44 \times 44\text{ cm}^3 \]
The volume of each spherical shot is:
\[ V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 2^3\text{ cm}^3 \]
Setting up the equation for total volume conservation:
\[ N \times V_{\text{shot}} = V_{\text{cube}} \]
\[ N = \frac{44 \times 44 \times 44}{\frac{4}{3} \times \frac{22}{7} \times 8} \]
\[ N = \frac{44 \times 44 \times 44 \times 3 \times 7}{88 \times 8} \]
\[ N = \frac{44 \times 44 \times 21}{16} = 11 \times 11 \times 21 = 2541 \]
Therefore, \( 2541 \) lead shots can be produced.
In simple words: Divide the volume of the large cube by the volume of one tiny spherical lead shot to get a total of 2541 shots.

Exam Tip: Keep numbers in factored form as long as possible. Canceling terms like 44 with 88 makes the arithmetic much faster and prevents multiplication mistakes.

Question. Three cubes of metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube of diagonal 24√3 cm. Find the edges of the three cubes.
Answer: Let the edges of the three original cubes be \( 3x \), \( 4x \), and \( 5x \). The sum of their volumes is:
\[ V_{\text{total}} = (3x)^3 + (4x)^3 + (5x)^3 = 27x^3 + 64x^3 + 125x^3 = 216x^3 \]
Let \( A \) be the edge length of the single large recast cube. The diagonal of a cube with side length \( A \) is \( A\sqrt{3} \). We are given:
\[ A\sqrt{3} = 24\sqrt{3}\text{ cm} \implies A = 24\text{ cm} \]
The volume of this new cube is:
\[ V_{\text{new}} = A^3 = 24 \times 24 \times 24\text{ cm}^3 \]
Since the volume is conserved during recasting:
\[ 216x^3 = 24 \times 24 \times 24 \]
\[ x^3 = \frac{24 \times 24 \times 24}{216} \]
Since \( 216 = 6^3 \), we get:
\[ x^3 = \left(\frac{24}{6}\right)^3 = 4^3 \implies x = 4 \]
The edges of the three cubes are:
- First edge: \( 3 \times 4 = 12\text{ cm} \)
- Second edge: \( 4 \times 4 = 16\text{ cm} \)
- Third edge: \( 5 \times 4 = 20\text{ cm} \)
In simple words: Find the size of the new cube from its diagonal, then set its volume equal to the sum of the volumes of the three smaller cubes to find their actual edges.

Exam Tip: Remember that the diagonal of a cube of side \( s \) is \( s\sqrt{3} \). Use this standard formula to instantly find the side of the recast cube without extra work.

Question. A heap of rice in the form of a cone of radius 3 m and height 3 m. Find the volume of the rice. How much cloth is required to just cover the heap?
Answer: For the conical heap of rice, the radius is \( r = 3\text{ m} \) and the height is \( h = 3\text{ m} \). The volume of the rice is:
\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 3^2 \times 3 = 9\pi\text{ m}^3 \]
To calculate the cloth required to cover the heap, we find the curved surface area of the cone. First, calculate the slant height \( l \):
\[ l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\text{ m} \]
The required area of the covering cloth is:
\[ \text{Cloth Area} = \pi r l = \pi \times 3 \times 3\sqrt{2} = 9\sqrt{2}\pi\text{ m}^2 \]
In simple words: Use the cone volume formula to find the heap's volume as \( 9\pi\text{ m}^3 \). The cloth area is equal to the curved surface area, which is \( 9\sqrt{2}\pi\text{ m}^2 \).

Exam Tip: Be sure to keep track of the units. When dimensions are given in meters, volume must be in \( \text{m}^3 \) and area must be in \( \text{m}^2 \).

Level-III

Question. Three solid metallic spheres of radii 3 cm , 4cm and 5cm respectively are melted to form a single solid sphere. Find the diameter of the resulting sphere.
Answer: Let the radii of the three metallic spheres be \( r_1 = 3\text{ cm} \), \( r_2 = 4\text{ cm} \), and \( r_3 = 5\text{ cm} \). Let \( R \) be the radius of the resulting sphere. The sum of the volumes of the three smaller spheres equals the volume of the single large sphere:
\[ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3 \]
\[ R^3 = r_1^3 + r_2^3 + r_3^3 \]
\[ R^3 = 3^3 + 4^3 + 5^3 \]
\[ R^3 = 27 + 64 + 125 = 216 \]
\[ R = \sqrt[3]{216} = 6\text{ cm} \]
The diameter \( D \) of the new sphere is:
\[ D = 2R = 2 \times 6 = 12\text{ cm} \]
In simple words: Add up the cubic volumes of the three smaller spheres to find the volume of the large sphere. This gives a radius of 6 cm, making the diameter 12 cm.

Exam Tip: Always double-check if the question asks for the radius or the diameter. A very common error is stopping at \( R = 6\text{ cm} \) and missing the final multiplication step.

Question. Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer: Each coin represents a small cylinder. Let \( n \) be the number of coins needed.
For one coin:
- Radius \( r = \frac{1.5}{2} = 0.75\text{ cm} = \frac{3}{4}\text{ cm} \)
- Thickness (height) \( h = 0.2\text{ cm} = \frac{1}{5}\text{ cm} \)
- Volume \( V_{\text{coin}} = \pi r^2 h = \pi \times \left(\frac{3}{4}\right)^2 \times \frac{1}{5} = \frac{9\pi}{80}\text{ cm}^3 \)
For the large cylinder:
- Radius \( R = \frac{4.5}{2} = 2.25\text{ cm} = \frac{9}{4}\text{ cm} \)
- Height \( H = 10\text{ cm} \)
- Volume \( V_{\text{cylinder}} = \pi R^2 H = \pi \times \left(\frac{9}{4}\right)^2 \times 10 = \frac{405\pi}{8}\text{ cm}^3 \)
Since total volume is conserved during recasting:
\[ n \times V_{\text{coin}} = V_{\text{cylinder}} \]
\[ n = \frac{V_{\text{cylinder}}}{V_{\text{coin}}} = \frac{\frac{405\pi}{8}}{\frac{9\pi}{80}} = \frac{405}{8} \times \frac{80}{9} = 45 \times 10 = 450 \]
Therefore, \( 450 \) coins must be melted.
In simple words: We treat the coins as cylinders and divide the large cylinder's volume by the volume of a single coin. This results in exactly 450 coins.

Exam Tip: Convert decimal values to fractions before performing calculations. Fractions are much easier to square and simplify, significantly reducing arithmetic mistakes.

Question. The rain water from a roof 22m x 20m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 of the vessel is just full . Find the rainfall in cm.
Answer: Let the depth of the rainfall on the roof be \( h\text{ m} \). The volume of water collected on the rectangular roof is:
\[ V_{\text{roof}} = 22 \times 20 \times h = 440h\text{ m}^3 \]
For the cylindrical vessel, the base diameter is \( 2\text{ m} \), so its radius is \( R = 1\text{ m} \). The height of the vessel is \( H = 3.5\text{ m} \). The volume of this cylinder is:
\[ V_{\text{cylinder}} = \pi R^2 H = \frac{22}{7} \times 1^2 \times 3.5 = 22 \times 0.5 = 11\text{ m}^3 \]
Since the drained rainwater completely fills the cylinder, we equate the two volumes:
\[ 440h = 11 \]
\[ h = \frac{11}{440} = \frac{1}{40}\text{ m} \]
To convert the rainfall depth from meters to centimeters, multiply by 100:
\[ h = \frac{1}{40} \times 100 = 2.5\text{ cm} \]
Hence, the rainfall depth is \( 2.5\text{ cm} \).
In simple words: The volume of water falling on the roof equals the volume of the cylindrical tank it fills. Equating these volumes gives us a rainfall depth of 2.5 cm.

Exam Tip: Pay close attention to unit conversions. The dimensions are given in meters, but the final answer must be converted and expressed in centimeters.

Question. The radius of the base and the height of a solid right circular cylinder are in the ratio of 2:3 and its volume is 1617 cm2. Find the total surface area of the cylinder.
Answer: Let the radius of the cylinder be \( r = 2x \) and the height be \( h = 3x \). The volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
We are given the volume is \( 1617\text{ cm}^3 \). Substituting the values:
\[ 1617 = \frac{22}{7} \times (2x)^2 \times (3x) \]
\[ 1617 = \frac{22}{7} \times 4x^2 \times 3x \]
\[ 1617 = \frac{264}{7}x^3 \]
\[ x^3 = \frac{1617 \times 7}{264} \]
Dividing by 33 on both sides:
\[ x^3 = \frac{49 \times 7}{8} = \frac{343}{8} \]
Taking the cube root:
\[ x = \frac{7}{2} = 3.5 \]
Therefore, the dimensions of the cylinder are:
- Radius \( r = 2 \times 3.5 = 7\text{ cm} \)
- Height \( h = 3 \times 3.5 = 10.5\text{ cm} \)
The total surface area (TSA) of the cylinder is:
\[ \text{TSA} = 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times (7 + 10.5) \]
\[ \text{TSA} = 44 \times 17.5 = 770\text{ cm}^2 \]
In simple words: We find the common ratio multiplier from the volume formula to get a radius of 7 cm and a height of 10.5 cm. Then, we find the total surface area, which is 770 sq. cm.

Exam Tip: Look out for perfect cubes like \( 343 = 7^3 \) and \( 8 = 2^3 \) during calculations. Recognizing them makes simplifying cubic equations very fast.

Question. A semispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are needed to empty the bowl?
Answer: Let \( R = 9\text{ cm} \) be the radius of the hemispherical bowl. The volume of liquid in the bowl is:
\[ V_{\text{bowl}} = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi \times 9^3 = 486\pi\text{ cm}^3 \]
For each small cylindrical bottle, the radius is \( r = \frac{3}{2} = 1.5\text{ cm} \) and the height is \( h = 4\text{ cm} \). The volume of one bottle is:
\[ V_{\text{bottle}} = \pi r^2 h = \pi \times 1.5^2 \times 4 = 9\pi\text{ cm}^3 \]
Let \( N \) be the number of bottles required:
\[ N = \frac{V_{\text{bowl}}}{V_{\text{bottle}}} = \frac{486\pi}{9\pi} = 54 \]
Therefore, \( 54 \) bottles are needed to empty the bowl.
In simple words: Divide the volume of liquid in the bowl by the volume of a single bottle. This shows that we need exactly 54 bottles.

Exam Tip: Leave both volume equations in terms of \( \pi \). Canceling \( \pi \) directly during division eliminates long calculations and saves time.

Level-IV

Question. A sphere of diameter 12 cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm. Find the diameter of the cylindrical vessel.
Answer: The diameter of the submerged sphere is \( 12\text{ cm} \), so its radius is \( R = 6\text{ cm} \). The volume of this sphere is:
\[ V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times 6^3 = 288\pi\text{ cm}^3 \]
Let \( r \) be the radius of the cylindrical vessel. The rise in water level is \( h = 3\frac{5}{9}\text{ cm} = \frac{32}{9}\text{ cm} \). The volume of displaced water that rises is:
\[ V_{\text{displaced}} = \pi r^2 h = \pi r^2 \times \frac{32}{9} \]
Since the volume of the displaced water must equal the volume of the sphere:
\[ \pi r^2 \times \frac{32}{9} = 288\pi \]
\[ r^2 \times \frac{32}{9} = 288 \]
\[ r^2 = \frac{288 \times 9}{32} \]
\[ r^2 = 9 \times 9 = 81 \implies r = 9\text{ cm} \]
The diameter of the cylinder is:
\[ D = 2r = 2 \times 9 = 18\text{ cm} \delta \]
Thus, the diameter of the cylindrical vessel is \( 18\text{ cm} \).
In simple words: The volume of water that rises in the cylindrical vessel is exactly equal to the volume of the dropped sphere. Solving this tells us the vessel's diameter is 18 cm.

Exam Tip: Remember that any submerged object displaces a volume of water equal to its own volume. Setting the cylinder volume rise equal to the sphere volume is the key to this question.

Question. A bucket made up of metal sheet in the form of a frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30cm and 10cm respectively. Find the cost of milk which can completely fill the bucket at rate of Rs 20 per litre and the cost of the metal sheet used, if it costs Rs 10 per 100 cm2. (use π = 3.14)
Answer: For the frustum-shaped bucket, the height is \( h = 24\text{ cm} \), the top radius is \( R = 15\text{ cm} \), and the bottom radius is \( r = 5\text{ cm} \). First, we calculate the volume of the bucket:
\[ V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \]
\[ V = \frac{1}{3} \times 3.14 \times 24 \times (15^2 + 5^2 + 15 \times 5) \]
\[ V = 3.14 \times 8 \times (225 + 25 + 75) = 25.12 \times 325 = 8164\text{ cm}^3 \]
Converting the volume to litres (since \( 1000\text{ cm}^3 = 1\text{ litre} \)):
\[ \text{Volume of milk} = \frac{8164}{1000} = 8.164\text{ litres} \]
At a rate of Rs. 20 per litre, the cost of milk is:
\[ \text{Cost} = 8.164 \times 20 = \text{Rs. } 163.28 \]
Next, we calculate the slant height \( l \) of the bucket:
\[ l = \sqrt{h^2 + (R - r)^2} = \sqrt{24^2 + (15 - 5)^2} = \sqrt{576 + 10^2} = \sqrt{676} = 26\text{ cm} \]
Since the bucket is open at the top, the surface area of the metal sheet used is the curved surface area plus the bottom base area:
\[ \text{Area} = \pi (R + r)l + \pi r^2 \]
\[ \text{Area} = 3.14 \times (15 + 5) \times 26 + 3.14 \times 5^2 \]
\[ \text{Area} = 3.14 \times 20 \times 26 + 3.14 \times 25 \]
\[ \text{Area} = 3.14 \times 520 + 3.14 \times 25 = 3.14 \times 545 = 1711.3\text{ cm}^2 \]
At a rate of Rs. 10 per \( 100\text{ cm}^2 \), the cost of the metal sheet is:
\[ \text{Cost} = \frac{1711.3}{100} \times 10 = \text{Rs. } 171.13 \]
In simple words: First, find the volume to calculate the capacity in litres, which costs Rs. 163.28. Then, calculate the total metal area (leaving the top open) to find the sheet cost of Rs. 171.13.

Exam Tip: A bucket has an open top. When calculating the total surface area of the metal sheet, only add the curved surface area and the bottom circle area \( (\pi r^2) \). Do not include the top circle area.

Question. A vessel is in the form of a semi spherical bowl mounted by a hollow cylinder. The diameter of the semisphere is 14cm and the total height of the vessel is 13 cm.Find the capacity of the vessel.(Take x=22/7]
Answer: The radius \( r \) of both the hemisphere and the cylinder is:
\[ r = \frac{14}{2} = 7\text{ cm} \delta \]
Since the total height of the vessel is \( 13\text{ cm} \) and the height of the hemispherical bowl is equal to its radius (\( 7\text{ cm} \)), the height \( h \) of the hollow cylinder is:
\[ h = 13 - 7 = 6\text{ cm} \]
The capacity (volume) of the vessel is the sum of the volumes of the cylinder and the hemisphere:
\[ V = V_{\text{cylinder}} + V_{\text{hemisphere}} \]
\[ V = \pi r^2 h + \frac{2}{3}\pi r^3 = \pi r^2 \left(h + \frac{2}{3}r\right) \]
Substituting \( r = 7\text{ cm} \) and \( h = 6\text{ cm} \):
\[ V = \frac{22}{7} \times 7^2 \times \left(6 + \frac{2}{3} \times 7\right) \]
\[ V = 154 \times \left(6 + \frac{14}{3}\right) = 154 \times \frac{32}{3} = \frac{4928}{3} \approx 1642.67\text{ cm}^3 \]
Thus, the capacity of the vessel is \( 1642.67\text{ cm}^3 \).
In simple words: Subtract the hemisphere's radius from the total height to find the cylinder's height (6 cm). Adding their volumes together gives a total capacity of 1642.67 cubic centimeters.

Exam Tip: A hemisphere's height is always equal to its radius. Use this fundamental property to split the total height of combined solids correctly.

Question. If the radii of the ends of a bucket, 45 cm height, are 28 cm and 7 cm determine the capacity and total suface area of the bucket,
Answer: The bucket is in the shape of a frustum of a cone with height \( h = 45\text{ cm} \), larger radius \( R = 28\text{ cm} \), and smaller radius \( r = 7\text{ cm} \). The capacity (volume) is:
\[ V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \]
\[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7) \]
\[ V = \frac{330}{7} \times (784 + 49 + 196) = \frac{330}{7} \times 1029 = 330 \times 147 = 48510\text{ cm}^3 \delta \]
To find the total surface area, we calculate the slant height \( l \) of the frustum:
\[ l = \sqrt{h^2 + (R - r)^2} = \sqrt{45^2 + (28 - 7)^2} = \sqrt{2025 + 21^2} = \sqrt{2466} \approx 49.66\text{ cm} \]
Since a bucket is open at the top, its total surface area includes the curved surface area and the base area:
\[ \text{Area} = \pi (R + r)l + \pi r^2 \]
\[ \text{Area} = \frac{22}{7} \times (28 + 7) \times 49.66 + \frac{22}{7} \times 7^2 \]
\[ \text{Area} = \frac{22}{7} \times 35 \times 49.66 + \frac{22}{7} \times 49 \]
\[ \text{Area} = 110 \times 49.66 + 154 = 5462.6 + 154 = 5616.6\text{ cm}^2 \]
In simple words: The volume of the bucket is calculated using the frustum formula, giving 48510 cubic centimeters. Then, we find the slant height (49.66 cm) to calculate the external surface area of the bucket, which is 5616.6 square centimeters.

Exam Tip: Simplify fractions step-by-step. Since \( 1029 \) is divisible by \( 7 \), performing this division first will keep your arithmetic clean and simple.

Question. A toy is in the form of a cone mounted on a hemisphere of common base radius 7cm. The total height of the toy is 13 cm . Find the total surface of the toy.( =22/7)
Answer: Referencing the geometric schematic below, we use the corrected total height of \( 31\text{ cm} \). The radius is \( r = 7\text{ cm} \). The height of the hemispherical base is equal to its radius (\( 7\text{ cm} \)). The height \( h \) of the conical top is:
\[ h = 31 - 7 = 24\text{ cm} \]
Now, we calculate the slant height \( l \) of the cone:
\[ l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm} \]
The total surface area of the toy is the sum of the curved surface area of the cone and the curved surface area of the hemisphere:
\[ \text{Total Area} = \pi r l + 2\pi r^2 = \pi r (l + 2r) \]
\[ \text{Total Area} = \frac{22}{7} \times 7 \times (25 + 2 \times 7) \]
\[ \text{Total Area} = 22 \times (25 + 14) = 22 \times 39 = 858\text{ cm}^2 \]
In simple words: The cone's height is found by subtracting the hemisphere's radius from the total height, giving 24 cm. Using this, we find the slant height is 25 cm, and the combined curved surface area is 858 square centimeters.

Exam Tip: When a solid is mounted on another, the flat circular interface is hidden inside. Make sure you only sum the curved surface areas of the individual shapes.

Self Evaluation Question

Question. The base radii of two right circular cone of the same height are in the ratio 3:5. Find the ratio of their volumes.
Answer: Let the common height of both cones be \( h \). Let their base radii be \( r_1 = 3x \) and \( r_2 = 5x \). The volume of a right circular cone is given by \( V = \frac{1}{3}\pi r^2 h \). The ratio of their volumes is:
\[ \frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h}{\frac{1}{3}\pi r_2^2 h} = \frac{r_1^2}{r_2^2} \]
Substituting the values of the radii:
\[ \frac{V_1}{V_2} = \frac{(3x)^2}{(5x)^2} = \frac{9x^2}{25x^2} = \frac{9}{25} \]
Thus, the ratio of their volumes is \( 9:25 \).
In simple words: Since their heights are identical, the ratio of their volumes is simply the square of the ratio of their base radii, which gives 9 to 25.

Exam Tip: For any two cones of equal height, the volume ratio is always equal to the square of their base radii: \( V_1:V_2 = r_1^2:r_2^2 \).

Question. If a,b,c are the dimensions of a cuboid, S be the total surface area and v its volume then prove that 1/v=2/s(1/a+1/b+1/c).
Answer: For a cuboid with dimensions \( a, b, c \), the volume \( v \) is given by:
\[ v = a \times b \times c = abc \]
The total surface area \( S \) is given by:
\[ S = 2(ab + bc + ca) \]
Let us begin with the right-hand side (R.H.S.) of the given expression:
\[ \text{R.H.S.} = \frac{2}{S}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \]
Taking the common denominator inside the parenthesis:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc} \]
Substituting this back into our expression:
\[ \text{R.H.S.} = \frac{2}{S} \left( \frac{ab + bc + ca}{abc} \right) \]
Since \( S = 2(ab + bc + ca) \), we can substitute \( ab + bc + ca = \frac{S}{2} \):
\[ \text{R.H.S.} = \frac{2}{S} \left( \frac{\frac{S}{2}}{abc} \right) \]
\[ \text{R.H.S.} = \frac{1}{abc} = \frac{1}{v} = \text{L.H.S.} \]
Hence proved.
In simple words: We find a common denominator for the sum of the reciprocal dimensions. Replacing the terms with surface area and volume simplifies the equation directly.

Exam Tip: In algebraic proof questions, it is usually much simpler to start with the more complex side (the R.H.S.) and simplify it to match the simpler side (the L.H.S.).

Question. If h, c, v respectively are the height, the curved surface area and volume of a cone prove that 3\pi vh3 -c2 h2 +9v2 =0
Answer: Let the radius of the cone's base be \( r \) and its slant height be \( l = \sqrt{r^2 + h^2} \). The curved surface area \( c \) is:
\[ c = \pi r l \implies c^2 = \pi^2 r^2 (r^2 + h^2) \]
The volume \( v \) of the cone is:
\[ v = \frac{1}{3}\pi r^2 h \implies r^2 = \frac{3v}{\pi h} \]
Substituting this expression for \( r^2 \) into the equation for \( c^2 \):
\[ c^2 = \pi^2 \left(\frac{3v}{\pi h}\right) \left(\frac{3v}{\pi h} + h^2\right) \]
\[ c^2 = \frac{3\pi v}{h} \left(\frac{3v + \pi h^3}{\pi h}\right) \]
\[ c^2 = \frac{3v(3v + \pi h^3)}{h^2} \]
Multiplying both sides by \( h^2 \):
\[ c^2 h^2 = 9v^2 + 3\pi v h^3 \]
Rearranging all terms to one side:
\[ 3\pi v h^3 - c^2 h^2 + 9v^2 = 0 \]
Hence, the relation is successfully proved.
In simple words: We express the radius squared in terms of volume and height, then substitute it into the curved surface area equation. Simplifying this gives us the desired identity.

Exam Tip: In cone proofs, expressing \( r^2 \) in terms of volume \( v \) and height \( h \) is a very powerful way to eliminate the radius term entirely.

Question. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. If the total height of the toy is 15.5 cm. Find the volume of the toy .(use \pi =22/7).
Answer: The radius of both the hemisphere and the conical part is \( r = 3.5\text{ cm} = \frac{7}{2}\text{ cm} \). The height of the hemispherical base is equal to its radius, \( 3.5\text{ cm} \). The height \( h \) of the conical top is:
\[ h = 15.5 - 3.5 = 12\text{ cm} \]
The total volume of the toy is the sum of the volumes of the hemisphere and the cone:
\[ V = V_{\text{hemisphere}} + V_{\text{cone}} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h \]
Factoring out the common terms:
\[ V = \frac{1}{3}\pi r^2 (2r + h) \]
\[ V = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times (2 \times 3.5 + 12) \]
\[ V = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times (7 + 12) \]
\[ V = \frac{11 \times 7}{6} \times 19 = \frac{1463}{6} \approx 243.83\text{ cm}^3 \]
Thus, the volume of the toy is approximately \( 243.83\text{ cm}^3 \).
In simple words: We subtract the hemisphere's radius from the total height to find the cone's height (12 cm). Then, we calculate and add the volumes of the hemisphere and the cone to get 243.83 cubic centimeters.

Exam Tip: Factoring out common algebraic terms at the start of the calculation speeds up the process and prevents arithmetic errors.