NCERT Class 10 Mathematics — Surface Areas and Volumes, Exercise 12.1. All 9 questions solved.

Every solid in this exercise is two or three familiar solids stuck together or cut out of one another, and the question is always the same: which surfaces can you actually touch? The formulae are the ones from Class 9, for radius $r$, height $h$ and slant height $l$:

$$\text{CSA of cylinder} = 2\pi r h \qquad \text{CSA of cone} = \pi r l,\ \ l = \sqrt{r^2 + h^2} \qquad \text{CSA of hemisphere} = 2\pi r^2$$

together with $6a^2$ for a cube and $2(lb + bh + hl)$ for a cuboid.

Key insight. Count faces, not solids. Where two solids are joined, the faces pressed together disappear from the surface, so the total surface area is not the sum of the two total surface areas. Before using any formula, list the surfaces a painter would have to cover: a hemisphere stuck on a flat face hides a disc of area $\pi r^2$ but adds a curved surface of $2\pi r^2$, and a hemisphere scooped out of a face does exactly the same.

Unless stated otherwise, take $\pi = \dfrac{22}{7}$.

Question 1

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

Solution. The edge of each cube is $a$ with $a^3 = 64$, so $a = 4$ cm. Placed end to end, the two cubes make a cuboid $8$ cm long, $4$ cm wide and $4$ cm high:

$$\text{surface area} = 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 160\ \text{cm}^2$$

The same answer from the key insight: two separate cubes have $2 \times 6 \times 16 = 192\ \text{cm}^2$ of surface, and joining them hides one face of each, $2 \times 16 = 32\ \text{cm}^2$, leaving $160\ \text{cm}^2$.

Question 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.

Solution. The radius of the hemisphere, and of the cylinder on it, is $r = 7$ cm. The hemisphere takes up $7$ cm of the total height, so the cylinder is $h = 13 – 7 = 6$ cm tall.

The inside of the vessel consists of the inner curved surface of the hemisphere and the inner curved surface of the cylinder. There is no lid, and the circle where the two parts meet is not a surface at all:

$$\text{inner surface area} = 2\pi r^2 + 2\pi r h = 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times (7 + 6) = 44 \times 13 = 572\ \text{cm}^2$$

Question 3

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.

Solution. The hemisphere accounts for $3.5$ cm of the height, so the cone’s vertical height is $h = 15.5 – 3.5 = 12$ cm. The cone’s curved surface needs the slant height:

$$l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\ \text{cm}$$

The flat base of the cone and the flat face of the hemisphere are glued together, so neither shows. What remains is the cone’s curved surface and the hemisphere’s curved surface:

$$\text{TSA} = \pi r l + 2\pi r^2 = \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 3.5^2 = 137.5 + 77 = 214.5\ \text{cm}^2$$

Question 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.

Solution. The hemisphere’s flat face must sit on the top face of the cube, a $7$ cm square. The largest circle that fits in that square has diameter equal to the side, so the greatest diameter is $7$ cm, and then $r = 3.5$ cm.

Now count surfaces. The cube contributes all six faces, except that a disc of area $\pi r^2$ on the top face is covered by the hemisphere. The hemisphere contributes its curved surface $2\pi r^2$:

$$\text{surface area} = 6a^2 – \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$$

$$= 6 \times 49 + \frac{22}{7} \times 3.5^2 = 294 + 38.5 = 332.5\ \text{cm}^2$$

Question 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.

Solution. Here $l$ is the edge of the cube, and the hemisphere has radius $\frac{l}{2}$. Cutting the depression removes a disc of radius $\frac{l}{2}$ from the top face, but exposes the curved inner surface of the hollow, which is the curved surface of a hemisphere:

$$\text{surface area} = 6l^2 – \pi\left(\frac{l}{2}\right)^2 + 2\pi\left(\frac{l}{2}\right)^2 = 6l^2 + \frac{\pi l^2}{4} = \frac{l^2}{4}(\pi + 24)$$

This is question 4 turned inside out: whether the hemisphere is added on top or dug out of the face, the surface gains $2\pi r^2 – \pi r^2 = \pi r^2$.

Question 6

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 $14$ mm and the diameter of the capsule is $5$ mm. Find its surface area.

Solution. The radius is $r = 2.5$ mm. Each hemispherical end takes up $2.5$ mm of the length, so the cylindrical part is $h = 14 – 2 \times 2.5 = 9$ mm long.

The outside of the capsule is the curved surface of the cylinder together with the curved surfaces of the two hemispheres, which between them make a whole sphere:

$$\text{surface area} = 2\pi r h + 2 \times 2\pi r^2 = 2\pi r(h + 2r)$$

$$= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = 2 \times \frac{22}{7} \times 2.5 \times 14 = 220\ \text{mm}^2$$

Question 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 ₹ $500$ per $\text{m}^2$. (Note that the base of the tent will not be covered with canvas.)

Solution. The radius is $r = 2$ m. Canvas covers the wall (the curved surface of the cylinder, $h = 2.1$ m) and the roof (the curved surface of the cone, $l = 2.8$ m). The floor is not covered, and the circle where roof meets wall is not a surface:

$$\text{canvas} = 2\pi r h + \pi r l = \pi r(2h + l) = \frac{22}{7} \times 2 \times (4.2 + 2.8) = \frac{22}{7} \times 2 \times 7 = 44\ \text{m}^2$$

$$\text{cost} = 44 \times 500 = ₹\,22000$$

The slant height is given directly, so the cone’s vertical height is never needed.

Question 8

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 $\text{cm}^2$.

Solution. Here $r = 0.7$ cm and $h = 2.4$ cm. The cone has the same base as the cylinder, so its base takes up the whole of one end, and its vertex is at the centre of the other end. Its slant height is

$$l = \sqrt{0.7^2 + 2.4^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\ \text{cm}$$

The remaining solid has three surfaces: the outer curved surface of the cylinder, the one flat end that is still intact, and the inner curved surface of the conical cavity. The other flat end has been removed entirely.

$$\text{TSA} = 2\pi r h + \pi r^2 + \pi r l = \pi r(2h + r + l)$$

$$= \frac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6\ \text{cm}^2$$

To the nearest $\text{cm}^2$, as the question asks, this is $18\ \text{cm}^2$.

Question 9

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 $10$ cm, and its base is of radius $3.5$ cm, find the total surface area of the article.

Solution. Each hemisphere has the same radius as the cylinder, $r = 3.5$ cm, so it removes the whole of that flat end and exposes a curved hollow in its place. (The two hollows are $3.5$ cm deep each, $7$ cm in all, less than the height of $10$ cm, so they do not meet.)

The surface of the article is the outer curved surface of the cylinder and the inner curved surfaces of the two hemispheres:

$$\text{TSA} = 2\pi r h + 2 \times 2\pi r^2 = 2\pi r(h + 2r)$$

$$= 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) = 22 \times 17 = 374\ \text{cm}^2$$

Common mistakes

• Question 1, adding the two cubes’ surface areas. That gives $192\ \text{cm}^2$ and counts the two faces pressed together, which are no longer on the outside of the cuboid.
• Question 2, using $13$ cm as the cylinder’s height. The $13$ cm is the whole vessel. The hemisphere below takes up $7$ cm of it, leaving $6$ cm for the cylinder.
• Question 3, using the vertical height in $\pi r l$. The curved surface of a cone needs the slant height, $12.5$ cm here, not the height $12$ cm, and certainly not the toy’s total height $15.5$ cm.
• Questions 4 and 5, forgetting the disc. Adding the hemisphere’s $2\pi r^2$ to all six faces of the cube counts the covered (or cut-away) disc of the top face as well. The net change is $+\pi r^2$ in both questions.
• Question 7, including the floor or the cone’s base. Canvas covers only the two curved surfaces. The note in the question rules out the floor, and the cone’s base is inside the tent.
• Question 8, counting both flat ends. The cavity’s base is the whole of one end, so that end has gone. Students who add $2\pi r^2$ and leave out the cavity’s inner surface get $2\pi r h + 2\pi r^2 \approx 13.6\ \text{cm}^2$, which is less than the correct $17.6$ — the hollow adds surface.

Practise next

• Exercise 12.2 — volumes of the same kinds of combined solids, where nothing disappears at a joint and the parts simply add.
• Chapter 11, Exercise 11.1 — the circle areas and circumferences that every one of these surface formulae is built on.