NCERT Class 10 Maths – Exercise 12.1 Solutions

NCERT Class 10 Maths

Chapter 12 – Surface Areas and Volumes | Exercise 12.1

(Rationalized Syllabus 2025-26)

💡 Concept: Surface Area of Combinations

When solids are joined together, the Total Surface Area (TSA) of the new solid is the sum of the Curved Surface Areas (CSA) of the visible parts. Faces that are glued together or hidden are not included.

Q1

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

Step 1: Find the edge of the cube
Volume of cube = $a^3 = 64$. So, $a = \sqrt[3]{64} = 4$ cm.
Step 2: Dimensions of Cuboid
When joined end to end, the length doubles, but width and height remain same:
Length ($l$) = $4 + 4 = 8$ cm
Breadth ($b$) = $4$ cm
Height ($h$) = $4$ cm
Step 3: Surface Area
$$TSA = 2(lb + bh + hl)$$
$$= 2(8\times4 + 4\times4 + 4\times8)$$
$$= 2(32 + 16 + 32) = 2(80) = 160$$
✔️ Surface Area = 160 cm²
Q2

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.

Dimensions:
Diameter = 14 cm $\Rightarrow$ Radius ($r$) = 7 cm.
Height of hemisphere = Radius = 7 cm.
Height of cylinder ($h$) = Total Height – Hemisphere Height = $13 – 7 = 6$ cm.
Inner Surface Area:
Area = CSA of Cylinder + CSA of Hemisphere
$$= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$$
$$= 2 \times \frac{22}{7} \times 7 (6 + 7)$$
$$= 44 \times 13 = 572$$
✔️ Inner Surface Area = 572 cm²
Q3

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

Dimensions:
Radius ($r$) = 3.5 cm = $7/2$ cm.
Height of cone ($h$) = $15.5 – 3.5 = 12$ cm.
Slant Height ($l$) of cone:
$$l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144}$$
$$= \sqrt{156.25} = 12.5 \text{ cm}$$
Total Surface Area:
TSA = CSA of Cone + CSA of Hemisphere
$$= \pi rl + 2\pi r^2 = \pi r(l + 2r)$$
$$= \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5)$$
$$= 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5$$
✔️ Total Surface Area = 214.5 cm²
Q4

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.

Part 1: Greatest Diameter
The hemisphere fits on top of the cube, so its maximum diameter equals the side of the cube.
✔️ Greatest Diameter = 7 cm
Part 2: Surface Area
The base of the hemisphere covers a circular part of the top face of the cube.
TSA = (TSA of Cube) – (Area of base of hemisphere) + (CSA of Hemisphere)
$$= 6a^2 – \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$$
Here $a=7$, $r=7/2$.
$$= 6(7)^2 + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$$
$$= 6(49) + \frac{77}{2} = 294 + 38.5 = 332.5$$
✔️ Surface Area = 332.5 cm²
Q5

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 = $l$. Radius of hemisphere $r = l/2$.
Similar to Q4, we add CSA of hemisphere and subtract the circular base.
TSA = (TSA of Cube) – (Base of Hemisphere circle) + (CSA of Hemisphere)
$$= 6l^2 – \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2$$
$$= 6l^2 + \pi \left(\frac{l}{2}\right)^2 = 6l^2 + \frac{\pi l^2}{4}$$
$$= \frac{1}{4}(24l^2 + \pi l^2) = \frac{l^2}{4}(24 + \pi)$$
✔️ Surface Area = $\frac{l^2}{4}(24 + \pi)$ sq units
Q6

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.

Dimensions:
Diameter = 5 mm $\Rightarrow r = 2.5$ mm.
Total length = 14 mm.
Length of cylinder ($h$) = Total – 2(Radius) = $14 – (2.5 + 2.5) = 9$ mm.
Surface Area:
Area = CSA of Cylinder + 2 $\times$ CSA of Hemisphere
$$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$$
$$= 2 \times \frac{22}{7} \times 2.5 (9 + 5)$$
$$= \frac{110}{7} \times 14 = 110 \times 2 = 220$$
✔️ Surface Area = 220 mm²
Q7

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 at the rate of ₹ 500 per m².

Dimensions:
Cylinder: $r = 2$ m, $h = 2.1$ m.
Cone: $r = 2$ m, slant height $l = 2.8$ m.
Area of Canvas:
Area = CSA of Cylinder + CSA of Cone (Base is not covered with canvas)
$$= 2\pi rh + \pi rl = \pi r(2h + l)$$
$$= \frac{22}{7} \times 2 (2 \times 2.1 + 2.8)$$
$$= \frac{44}{7} (4.2 + 2.8) = \frac{44}{7} \times 7 = 44 \text{ m}^2$$
Cost:
$$\text{Cost} = 44 \times 500 = 22000$$
✔️ Area = 44 m² | Cost = ₹ 22,000
Q8

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

Radius $r = 0.7$ cm. Height $h = 2.4$ cm.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{0.7^2 + 2.4^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5$ cm.
Surface Area:
We have CSA of Cylinder + Area of Circular Base (Top is open) + CSA of Cone (Inside).
$$= 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$$
$$= \frac{22}{7} \times 0.7 (2 \times 2.4 + 0.7 + 2.5)$$
$$= 2.2 (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6$$
To nearest cm²:
✔️ Surface Area $\approx$ 18 cm²
Q9

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Radius $r = 3.5$ cm. Height $h = 10$ cm.
TSA = CSA Cylinder + 2 $\times$ CSA Hemisphere
$$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$$
$$= 2 \times \frac{22}{7} \times 3.5 (10 + 7)$$
$$= 22 \times 17 = 374$$
✔️ Surface Area = 374 cm²
🎉 Exercise 12.1 Completed | Chapter 12 Finished!
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