---
title: "NCERT Solutions Class 10 Maths Ch 12 Surface Areas and Volumes"
url: https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions
dateModified: 2026-10-07T15:55:53+00:00
---

# NCERT Solutions Class 10 Maths Ch 12 Surface Areas and Volumes

This chapter's questions cover the calculation of surface areas and volumes for various solid objects formed by combining basic shapes like cuboids, cones, cylinders, spheres, and hemispheres.

Free PDF (8 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-12-surface-areas-and-volumes-1cfc67625e.pdf

## EXERCISE 12.1

### Question 1

*3 marks · Short answer*

2 cubes each of volume $64\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

**Solution**

1. Let the side of each cube be $a$.
2. Volume of a cube $= a^3 = 64\text{ cm}^3$, which gives $a = 4\text{ cm}$.
3. When two cubes are joined end to end, the dimensions of the resulting cuboid are length $l = 4 + 4 = 8\text{ cm}$, breadth $b = 4\text{ cm}$, and height $h = 4\text{ cm}$.
4. Surface area of the resulting cuboid $= 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$.

**Answer:** $160\text{ cm}^2$

> Common mistake: Adding the total surface areas of the two cubes directly instead of finding the dimensions of the new cuboid.

### Question 2

*3 marks · Short answer*

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel.

**Solution**

1. Radius of the hemisphere $r = \frac{14}{2} = 7\text{ cm}$, which is also the radius of the cylinder.
2. Height of the cylinder $h = \text{Total height} - \text{Radius of hemisphere} = 13 - 7 = 6\text{ cm}$.
3. Inner surface area of the vessel $=$ CSA of cylinder $+$ CSA of hemisphere $= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$.
4. Substituting the values: $2 \times \frac{22}{7} \times 7 \times (6 + 7) = 44 \times 13 = 572\text{ cm}^2$.

**Answer:** $572\text{ cm}^2$

> Common mistake: Taking the height of the cylinder as the total height of the vessel ($13\text{ cm}$).

### Question 3

*3 marks · Short answer*

A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy.

**Solution**

1. Radius of the hemisphere and the cone $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
2. Height of the cone $h = \text{Total height} - \text{Radius} = 15.5 - 3.5 = 12\text{ cm}$.
3. Slant height of the cone $l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}$.
4. Total surface area of the toy $=$ CSA of hemisphere $+$ CSA of cone $= 2\pi r^2 + \pi rl = \pi r(2r + l)$.
5. Substituting the values: $\frac{22}{7} \times 3.5 \times (2(3.5) + 12.5) = 11 \times (7 + 12.5) = 11 \times 19.5 = 214.5\text{ cm}^2$.

**Answer:** $214.5\text{ cm}^2$

> Common mistake: Using the total height of the toy as the height of the cone.

### Question 4

*3 marks · Short answer*

A cubical block of side $7\text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

**Solution**

1. The greatest diameter that the hemisphere can have is equal to the edge of the cube, which is $7\text{ cm}$.
2. Radius of the hemisphere $r = \frac{7}{2}\text{ cm}$.
3. Total surface area of the solid $=$ TSA of cube $-$ base area of hemisphere $+$ CSA of hemisphere $= 6 \times \text{edge}^2 - \pi r^2 + 2\pi r^2 = 6 \times \text{edge}^2 + \pi r^2$.
4. Substituting the values: $6 \times 7^2 + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 6 \times 49 + 38.5 = 294 + 38.5 = 332.5\text{ cm}^2$.

**Answer:** $332.5\text{ cm}^2$

> Common mistake: Adding the total surface area of the hemisphere without subtracting its circular base area from the cube surface.

### Question 5

*3 marks · Short answer*

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

**Solution**

1. Let the edge of the cube be $l$. The diameter of the hemisphere is $l$, so its radius $r = \frac{l}{2}$.
2. Total surface area of the remaining solid $=$ TSA of cube $-$ base area of hemisphere $+$ CSA of hemisphere.
3. TSA of solid $= 6 \times \text{edge}^2 - \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2$.
4. Substituting the radius: $6l^2 + \pi \left(\frac{l}{2}\right)^2 = 6l^2 + \frac{1}{4}\pi l^2 = \frac{1}{4}l^2(24 + \pi)\text{ sq. units}$.

**Answer:** $\frac{1}{4}l^2(24 + \pi)\text{ sq. units}$

> Common mistake: Forgetting to add the curved surface area of the depression inside the block.

### Question 6

*3 marks · Short answer*

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area.

**Solution**

1. Diameter of the capsule $= 5\text{ mm}$, so radius of the cylindrical part and hemispheres $r = \frac{5}{2}\text{ mm} = 2.5\text{ mm}$.
2. Length of the cylindrical part $h = \text{Total length} - \text{Radius of left hemisphere} - \text{Radius of right hemisphere} = 14 - 2.5 - 2.5 = 9\text{ mm}$.
3. Total surface area of the capsule $=$ CSA of cylinder $+$ CSA of two hemispheres $= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
4. Substituting the values: $2 \times \frac{22}{7} \times \frac{5}{2} \times (9 + 2 \times 2.5) = \frac{110}{7} \times 14 = 220\text{ mm}^2$.

**Answer:** $220\text{ mm}^2$

> Common mistake: Taking the height of the cylinder as the total length of the capsule ($14\text{ mm}$).

### Question 7

*3 marks · Short answer*

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ $500\text{ per m}^2$. (Note that the base of the tent will not be covered with canvas.)

**Solution**

1. Given height of cylinder $h = 2.1\text{ m}$, diameter $= 4\text{ m}$, so radius $r = 2\text{ m}$, and slant height of cone $l = 2.8\text{ m}$.
2. Area of canvas used = CSA of cylinder + CSA of cone $= 2\pi rh + \pi rl = \pi r(2h + l)$.
3. Substituting the values, Area $= \frac{22}{7} \times 2 \times (2(2.1) + 2.8) = \frac{44}{7} \times (4.2 + 2.8) = \frac{44}{7} \times 7 = 44\text{ m}^2$.
4. Cost of canvas at the rate of ₹ $500\text{ per m}^2 = 44 \times 500 = \text{₹ } 22000$.

**Answer:** Area of the canvas is $44\text{ m}^2$ and the cost is ₹ $22000$.

> Common mistake: Including the base area of the cylinder in the total surface area calculation.

### Question 8

*3 marks · Short answer*

From a solid cylinder whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\text{cm}^2$.

**Solution**

1. Given height of cylinder $h = 2.4\text{ cm}$, diameter $= 1.4\text{ cm}$, so radius $r = 0.7\text{ cm}$.
2. Slant height of the conical cavity $l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$.
3. Total surface area of the remaining solid = CSA of cylinder + CSA of cone + base area of cylinder $= 2\pi rh + \pi rl + \pi r^2 = \pi r(2h + l + r)$.
4. Substituting the values, TSA $= \frac{22}{7} \times 0.7 \times (2(2.4) + 2.5 + 0.7) = 2.2 \times (4.8 + 3.2) = 2.2 \times 8 = 17.6\text{ cm}^2$, which is approximately $18\text{ cm}^2$ to the nearest $\text{cm}^2$.

**Answer:** $18\text{ cm}^2$

> Common mistake: Forgetting to add the base area of the cylinder or the curved surface area of the inner cone.

### Question 9

*3 marks · Short answer*

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is $10\text{ cm}$, and its base is of radius $3.5\text{ cm}$, find the total surface area of the article.

**Solution**

1. Given height of cylinder $h = 10\text{ cm}$ and radius of base $r = 3.5\text{ cm}$.
2. Total surface area of the article = CSA of cylinder + CSA of two hemispheres $= 2\pi rh + 2 \times (2\pi r^2) = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)$.
3. Substituting the values, TSA $= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2(3.5)) = 22 \times (10 + 7) = 22 \times 17 = 374\text{ cm}^2$.

**Answer:** $374\text{ cm}^2$

> Common mistake: Subtracting the area of the hemisphere instead of adding its curved surface area.

## EXERCISE 12.2

### Question 1

*3 marks · Short answer*

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

**Solution**

1. Radius of hemisphere $r = 1\text{ cm}$, radius of cone $r = 1\text{ cm}$, and height of cone $h = 1\text{ cm}$.
2. Volume of the solid = Volume of the hemisphere + Volume of the cone = $\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h$.
3. Volume = $\frac{2}{3}\pi (1)^3 + \frac{1}{3}\pi (1)^2 (1) = \frac{2}{3}\pi + \frac{1}{3}\pi = \pi\text{ cm}^3$.

**Answer:** $\pi\text{ cm}^3$

> Common mistake: Forgetting to add the two volumes or substituting incorrect values for height and radius.

### Question 2

*3 marks · Short answer*

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is $3\text{ cm}$ and its length is $12\text{ cm}$. If each cone has a height of $2\text{ cm}$, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

**Solution**

1. Given diameter $= 3\text{ cm}$, so radius $r = 1.5\text{ cm}$, length of model $= 12\text{ cm}$, and height of each cone $= 2\text{ cm}$.
2. Height of the cylindrical part $h = 12 - 2 - 2 = 8\text{ cm}$.
3. Volume of air in the model = Volume of cylinder + $2 \times$ Volume of cone $= \pi r^2 h + 2 \times \left(\frac{1}{3} \pi r^2 h_{\text{cone}}\right) = \pi r^2 \left(h + \frac{2}{3} h_{\text{cone}}\right)$.
4. Substituting the values: $\frac{22}{7} \times (1.5)^2 \times \left(8 + \frac{2}{3} \times 2\right) = \frac{22}{7} \times 2.25 \times \left(8 + \frac{4}{3}\right) = \frac{22}{7} \times 2.25 \times \frac{28}{3} = 66\text{ cm}^3$.

**Answer:** $66\text{ cm}^3$

> Common mistake: Taking the height of the cylinder equal to the total length of the model without subtracting the heights of the two cones.

### Question 3

*3 marks · Short answer*

A gulab jamun, contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $5\text{ cm}$ and diameter $2.8\text{ cm}$ (see Fig. 12.15).

**Solution**

1. Radius of each hemispherical end $r = \frac{2.8}{2} = 1.4\text{ cm}$.
2. Length of the cylindrical part $h = 5 - 2(1.4) = 5 - 2.8 = 2.2\text{ cm}$.
3. Volume of one gulab jamun = Volume of cylinder + $2 \times$ Volume of hemisphere = $\pi r^2 h + \frac{4}{3} \pi r^3 = \pi r^2 \left(h + \frac{2}{3}r\right)$.
4. Volume of one gulab jamun = $\frac{22}{7} \times 1.4 \times 1.4 \times \left(2.2 + \frac{2}{3} \times 1.4\right) = 6.16 \times \left(2.2 + \frac{2.8}{3}\right) = 6.16 \times \frac{9.4}{3} = 25.0493\text{ cm}^3$.
5. Volume of $45$ gulab jamuns = $45 \times 25.0493 = 1127.22\text{ cm}^3$.
6. Amount of sugar syrup = $30\%$ of total volume = $0.30 \times 1127.22 = 338.166\text{ cm}^3 \approx 338\text{ cm}^3$.

**Answer:** $338\text{ cm}^3$ (or $338.18\text{ cm}^3$ depending on $\pi$ approximation)

> Common mistake: Taking the height of the cylinder as the total length of $5\text{ cm}$ without subtracting the radii of the two hemispheres.

### Question 4

*3 marks · Short answer*

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand (see Fig. 12.16).

**Solution**

1. Dimensions of cuboid are $l = 15\text{ cm}$, $b = 10\text{ cm}$, $h = 3.5\text{ cm}$. Volume of cuboid = $15 \times 10 \times 3.5 = 525\text{ cm}^3$.
2. Radius of each conical depression $r = 0.5\text{ cm}$, depth $h = 1.4\text{ cm}$. Volume of one cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = \frac{11}{30}\text{ cm}^3$.
3. Volume of wood = Volume of cuboid $-$ 4 $\times$ Volume of one cone = $525 - 4 \times \frac{11}{30} = 525 - 1.47 = 523.53\text{ cm}^3$.

**Answer:** $523.53\text{ cm}^3$

> Common mistake: Adding the volume of the cones instead of subtracting them from the cuboid volume.

### Question 5

*3 marks · Short answer*

A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top, which is open, is $5\text{ cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

**Solution**

1. Given height of cone $h = 8\text{ cm}$ and radius $r = 5\text{ cm}$.
2. Volume of water in the conical vessel = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5)^2 (8) = \frac{200}{3} \pi\text{ cm}^3$.
3. Volume of one spherical lead shot of radius $r_1 = 0.5\text{ cm}$ = $\frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \pi \left(\frac{1}{2}\right)^3 = \frac{1}{6} \pi\text{ cm}^3$.
4. Volume of water that flows out = $\frac{1}{4} \times \text{Volume of cone} = \frac{1}{4} \times \frac{200}{3} \pi = \frac{50}{3} \pi\text{ cm}^3$.
5. Number of lead shots = $\frac{\text{Volume of water that flows out}}{\text{Volume of one lead shot}} = \frac{\frac{50}{3} \pi}{\frac{1}{6} \pi} = \frac{50}{3} \times 6 = 100$.

**Answer:** $100$

> Common mistake: Forgetting to multiply by $\frac{1}{4}$ for the water that flows out.

### Question 6

*3 marks · Short answer*

A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{g}$ mass. (Use $\pi = 3.14$)

**Solution**

1. Given first cylinder height $h_1 = 220\text{ cm}$, radius $r_1 = \frac{24}{2} = 12\text{ cm}$.
2. Given second cylinder height $h_2 = 60\text{ cm}$, radius $r_2 = 8\text{ cm}$.
3. Volume of the first cylinder = $\pi r_1^2 h_1 = 3.14 \times (12)^2 \times 220 = 3.14 \times 144 \times 220 = 99475.2\text{ cm}^3$.
4. Volume of the second cylinder = $\pi r_2^2 h_2 = 3.14 \times (8)^2 \times 60 = 3.14 \times 64 \times 60 = 12057.6\text{ cm}^3$.
5. Total volume of the pole = $99475.2 + 12057.6 = 111532.8\text{ cm}^3$.
6. Mass of the pole = $111532.8 \times 8\text{ g} = 892262.4\text{ g} = 892.26\text{ kg}$ (or approx. $892.27\text{ kg}$ depending on $\pi$).

**Answer:** $892.26\text{ kg}$

> Common mistake: Using diameter instead of radius for the first cylinder.

### Question 7

*3 marks · Short answer*

A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$.

**Solution**

1. Given cone height $h = 120\text{ cm}$, radius $r = 60\text{ cm}$, and hemisphere radius $r = 60\text{ cm}$.
2. Cylinder radius $R = 60\text{ cm}$ and height $H = 180\text{ cm}$.
3. Volume of the solid = Volume of cone + Volume of hemisphere = $\frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{1}{3} \pi (60)^2 (120) + \frac{2}{3} \pi (60)^3$.
4. Volume of solid = $\frac{1}{3} \pi (3600)(120) + \frac{2}{3} \pi (216000) = 144000\pi + 144000\pi = 288000\pi\text{ cm}^3$.
5. Volume of the cylinder = $\pi R^2 H = \pi (60)^2 (180) = 648000\pi\text{ cm}^3$.
6. Volume of water left = Volume of cylinder $-$ Volume of solid = $648000\pi - 288000\pi = 360000\pi\text{ cm}^3$.
7. Volume of water left = $360000 \times \frac{22}{7} = \frac{7920000}{7}\text{ cm}^3 = 1131428.57\text{ cm}^3 = 1.131\text{ m}^3$.

**Answer:** $1.13\text{ m}^3$ (or $1131428.57\text{ cm}^3$)

> Common mistake: Subtracting the height of the cone and hemisphere incorrectly from the cylinder height.

### Question 8

*3 marks · Short answer*

A spherical glass vessel has a cylindrical neck $8\text{ cm}$ long, $2\text{ cm}$ in diameter; the diameter of the spherical part is $8.5\text{ cm}$. By measuring the amount of water it holds, a child finds its volume to be $345\text{ cm}^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.

**Solution**

1. Given cylindrical neck length $h = 8\text{ cm}$, diameter $d = 2\text{ cm}$ (radius $r = 1\text{ cm}$).
2. Diameter of spherical part $d_1 = 8.5\text{ cm}$ (radius $r_1 = 4.25\text{ cm}$).
3. Volume of the cylindrical neck = $\pi r^2 h = 3.14 \times (1)^2 \times 8 = 25.12\text{ cm}^3$.
4. Volume of the spherical part = $\frac{4}{3} \pi r_1^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 = \frac{4}{3} \times 3.14 \times 76.765625 = 321.39\text{ cm}^3$.
5. Total volume of the vessel = $25.12 + 321.39 = 346.51\text{ cm}^3$.
6. Since $346.51\text{ cm}^3 \neq 345\text{ cm}^3$, the child is incorrect.

**Answer:** She is incorrect; the correct volume is $346.51\text{ cm}^3$.

> Common mistake: Using radius as 8.5 cm instead of 4.25 cm for the spherical part.

## Frequently asked questions

### How many exercises and questions are there in NCERT Class 10 Maths Chapter 12 Surface Areas and Volumes?

This chapter for the 2026-27 session contains two main exercises with a total of 17 questions. Exercise 12.1 has 9 questions focusing on surface areas, while Exercise 12.2 has 8 questions dealing with volumes of combined solids. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### What topics do the questions cover in Chapter 12 of Class 10 Maths?

The questions cover surface areas and volumes of various combined solids like cylinders, cones, hemispheres, spheres, cubes, and cuboids. Specific problems involve finding the surface area of tents, capsules, toys, and vessels, as well as calculating the volume and mass of combined shapes and conical cavities. SwaVid provides detailed answers to all these concepts on this page.

### Which are considered the hardest question types in this chapter and how should I approach them?

Questions involving complex combinations such as a cylinder with two conical ends, a conical cavity hollowed out from a solid cylinder, or vessels made of a cylinder and hemisphere are often tricky. To approach them, break the solid down into its basic shapes, identify which surfaces or volumes need to be added or subtracted, and apply the formulas carefully. SwaVid's step-by-step solutions on this page make these difficult problems much easier to understand.

### How can I write answers for full marks in Class 10 Maths Chapter 12 board exams?

To secure full marks, always start your answer by writing the given values and drawing a neat, labeled diagram of the combined solid. Clearly state the formulas used, substitute the values with proper units like $\text{cm}^2$ or $\text{cm}^3$, and show your calculation steps clearly. You can study the solved examples in SwaVid's free PDF available on this page to learn the correct presentation format.

### Is a free PDF of NCERT Solutions for Class 10 Maths Chapter 12 available?

Yes, a comprehensive and free PDF containing complete NCERT solutions for this chapter is available right here on this page. These solutions are prepared according to the 2026-27 syllabus to help you revise concepts and practice effectively for your exams.

## Related pages

- [Exercise 12.1 solutions](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-1)
- [Exercise 12.2 solutions](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-2)
- [Surface Areas and Volumes: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/surface-areas-and-volumes)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student