---
title: "NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1"
url: https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-1
dateModified: 2026-10-07T15:55:53+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Chapter 12: Surface Areas and Volumes. Every question from Exercise 12.1, with full working and the final answer.

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.

## Related pages

- [All Chapter 12 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions)
- [Exercise 12.2](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-2)

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