Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 12 Exercise 12.1 on calculating the Surface Areas of Combinations of Solids. Master finding the surface area of composite shapes like a cuboid formed by joining two cubes (Q.1), a vessel combining a cylinder and hemisphere (Q.2), and a toy combining a cone and hemisphere (Q.3). Learn how to determine the surface area of objects with hollowed-out sections, such as the cubical block with a hemispherical depression (Q.5) and the cylinder with a conical cavity (Q.8). Solutions emphasize finding the Curved Surface Area (CSA) of each component to calculate the total exposed area, including complex examples like the medicine capsule (Q.6) and the canvas for a tent (Q.7). Essential practice for composite 3D geometry problems.

Rbse Solutions for Class 10 Maths Exercise 12.1

Rbse Solutions for Class 10 Maths Exercise 12.2

Rbse Solutions for Class 10 Maths Chapter 12 Exercise 12.1 | Surface Areas and Volumes

Unless stated otherwise, use $\pi = 22/7$.

1. Surface Area of Resulting Cuboid

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Find the edge of the cube ($a$):$$V_{\text{cube}} = a^3 \implies 64 = a^3 \implies a = 4 \text{ cm}$$
Dimensions of the resulting cuboid ($l, b, h$):When two cubes are joined end-to-end, the length doubles.$$l = 4 + 4 = 8 \text{ cm}, \quad b = 4 \text{ cm}, \quad h = 4 \text{ cm}$$
Calculate the Surface Area (SA):$$\text{SA} = 2(lb + bh + hl)$$$$\text{SA} = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 2(80) = 160 \text{ cm}^2$$The surface area of the resulting cuboid is $\mathbf{160 \text{ cm}^2}$.

2. Inner Surface Area of Vessel

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

The vessel is a hollow cylinder mounted on a hollow hemisphere.

Identify Dimensions:
- Diameter $= 14 \text{ cm} \implies$ Radius ($r$): $7 \text{ cm}$.
- Total height $= 13 \text{ cm}$.
- Height of cylinder ($h$): $13 – r = 13 – 7 = 6 \text{ cm}$.
Calculate Inner Surface Area (ISA):$$\text{ISA} = \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{hemisphere}}$$$$\text{ISA} = 2\pi r h + 2\pi r^2 = 2\pi r (h + r)$$$$\text{ISA} = 2 \times \frac{22}{7} \times 7 \times (6 + 7) = 44 \times 13 = 572 \text{ cm}^2$$The inner surface area of the vessel is $\mathbf{572 \text{ cm}^2}$.

3. Total Surface Area of Toy

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

The toy is a cone mounted on a hemisphere.

Identify Dimensions:
- Radius ($r$): $3.5 \text{ cm}$ (or $7/2 \text{ cm}$).
- Total height $= 15.5 \text{ cm}$.
- Height of cone ($h$): $15.5 – 3.5 = 12 \text{ cm}$.
Find Slant Height ($l$):$$l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}$$
Calculate Total Surface Area (TSA):$$\text{TSA} = \text{CSA}_{\text{cone}} + \text{CSA}_{\text{hemisphere}} = \pi r l + 2\pi r^2 = \pi r (l + 2r)$$$$\text{TSA} = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5)$$$$\text{TSA} = 22 \times 0.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2$$The total surface area of the toy is $\mathbf{214.5 \text{ cm}^2}$.

4. Surface Area of Cubical Block Surmounted by Hemisphere

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

Greatest Diameter:The greatest diameter the hemisphere can have is equal to the edge of the cube.$$\text{Greatest Diameter} = \mathbf{7 \text{ cm}}$$$$\text{Radius ($r$)} = 7/2 = 3.5 \text{ cm}$$
Calculate Surface Area (SA) of the solid:The base area of the hemisphere covers the base area on the cube’s face, so we add the Curved Surface Area of the hemisphere to the Cube’s TSA, and subtract the circular base that’s hidden.$$\text{SA} = \text{TSA}_{\text{cube}} – \text{Area}_{\text{base of hemisphere}} + \text{CSA}_{\text{hemisphere}}$$$$\text{SA} = 6a^2 + \pi r^2$$$$\text{SA} = 6(7)^2 + \frac{22}{7} \times (3.5)^2$$$$\text{SA} = 294 + \frac{22}{7} \times 12.25 = 294 + 38.5 = 332.5 \text{ cm}^2$$The surface area of the solid is $\mathbf{332.5 \text{ cm}^2}$.

5. Surface Area of Remaining Solid (Hemispherical Depression)

5. 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.

Edge of cube ($a$): $l$.
Diameter of hemisphere: $l \implies$ Radius ($r$): $l/2$.

The depression means the flat circular area is removed, but the curved interior surface is exposed, so the calculation is the same as the previous problem:

$$\text{SA} = \text{TSA}_{\text{cube}} – \text{Area}_{\text{base of hemisphere}} + \text{CSA}_{\text{hemisphere}}$$

$$\text{SA} = 6a^2 + \pi r^2$$

Substitute $a = l$ and $r = l/2$:

$$\text{SA} = 6l^2 + \pi \left(\frac{l}{2}\right)^2 = 6l^2 + \frac{\pi l^2}{4}$$

$$\text{SA} = \mathbf{l^2 \left(6 + \frac{\pi}{4}\right) \text{ unit}^2}$$

6. Surface Area of Medicine Capsule

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.

The capsule is a cylinder with two hemispheres stuck to the ends.

Identify Dimensions:
- Total length ($L$): $14 \text{ mm}$.
- Diameter $= 5 \text{ mm} \implies$ Radius ($r$): $2.5 \text{ mm}$.
- Height of cylinder ($h$): $L – 2r = 14 – (2 \times 2.5) = 14 – 5 = 9 \text{ mm}$.
Calculate Surface Area (TSA):$$\text{TSA} = \text{CSA}_{\text{cylinder}} + 2 \times \text{CSA}_{\text{hemisphere}}$$$$\text{TSA} = 2\pi r h + 2 \times (2\pi r^2) = 2\pi r (h + 2r)$$$$\text{TSA} = 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = 2 \times \frac{22}{7} \times 2.5 \times 14$$$$\text{TSA} = 2 \times 22 \times 2.5 \times 2 = 220 \text{ mm}^2$$The surface area of the capsule is $\mathbf{220 \text{ mm}^2}$.

7. Area and Cost of Canvas for Tent

7. 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 m and 4 m, respectively, and the slant height of the top is 2.8 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 Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)

The tent is a cylinder surmounted by a cone. The base is not covered.

Identify Dimensions:
- Radius ($r$): $4/2 = 2 \text{ m}$.
- Height of cylinder ($h$): $2.1 \text{ m}$.
- Slant height of cone ($l$): $2.8 \text{ m}$.
Calculate Area of Canvas:$$\text{Area} = \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}} = 2\pi r h + \pi r l = \pi r (2h + l)$$$$\text{Area} = \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8) = \frac{44}{7} \times (4.2 + 2.8) = \frac{44}{7} \times 7 = 44 \text{ m}^2$$The area of the canvas is $\mathbf{44 \text{ m}^2}$.
Calculate Cost:$$\text{Cost} = \text{Area} \times \text{Rate} = 44 \times 500 = ₹ 22,000$$The cost of the canvas is $₹ 22,000$.

8. Total Surface Area of Remaining Solid

8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 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 cm².

A conical cavity is hollowed out from a solid cylinder.

Identify Dimensions:
- Height ($h$): $2.4 \text{ cm}$.
- Diameter $1.4 \text{ cm} \implies$ Radius ($r$): $0.7 \text{ cm}$.
Find Slant Height ($l$):$$l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5 \text{ cm}$$
Calculate Total Surface Area (TSA):The TSA of the remaining solid includes the bottom circular base of the cylinder, the curved surface of the cylinder, and the curved surface of the conical cavity.$$\text{TSA} = \text{Area}_{\text{base}} + \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}}$$$$\text{TSA} = \pi r^2 + 2\pi r h + \pi r l = \pi r (r + 2h + l)$$$$\text{TSA} = \frac{22}{7} \times 0.7 \times (0.7 + 2 \times 2.4 + 2.5)$$$$\text{TSA} = 2.2 \times (0.7 + 4.8 + 2.5) = 2.2 \times 8.0 = 17.6 \text{ cm}^2$$To the nearest $\text{cm}^2$, the total surface area is $\mathbf{18 \text{ cm}^2}$.

9. Total Surface Area of Wooden Article

The article is a solid cylinder with a hemisphere scooped out from each end.

Identify Dimensions:
- Height of cylinder ($h$): $10 \text{ cm}$.
- Radius ($r$): $3.5 \text{ cm}$.
Calculate Total Surface Area (TSA):The TSA is the sum of the CSA of the cylinder and the CSA of the two scooped-out hemispheres.$$\text{TSA} = \text{CSA}_{\text{cylinder}} + 2 \times \text{CSA}_{\text{hemisphere}}$$$$\text{TSA} = 2\pi r h + 2 \times (2\pi r^2) = 2\pi r (h + 2r)$$$$\text{TSA} = 2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)$$$$\text{TSA} = 2 \times 22 \times 0.5 \times (10 + 7) = 22 \times 17 = 374 \text{ cm}^2$$The total surface area of the article is $\mathbf{374 \text{ cm}^2}$.