CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 07

Surface Area and Volume of Solid Figures

In Class IX, we learnt how to find the surface areas and volumes of some regular solids like cuboid, cube, cylinder, cone and sphere. In our day to day life we come across various solids which are combinations of two or more such solids. For example, a conical circus tent with cylindrical base is a combination of a right circular cylinder and a right circular cone. In this chapter, we shall learn how to find the surface areas and volumes of such combinations.

(a) Cuboid

Total Surface Area (T.S.A.): The area of the surface from which the cuboid is formed. There are six faces (rectangular), eight vertices and twelve edges in a cuboid.

1. Total Surface Area (T.S.A.) = 2 [ℓ × b + b × h + h × ℓ]
2. Lateral Surface Area (L.S.A.), or area of 4 walls = 2h [ℓ + b]
3. Volume of cuboid = (Area of base) × height = (ℓ × b) × h
4. Length of diagonal = \( \sqrt{\ell^2 + b^2 + h^2} \)

(b) Cube

A cube has six faces. Each face is a square.

1. T.S.A. = 2 [x·x + x·x + x·x] = 2 [x² + x² + x²] = 2 (3x²) = 6x²
2. L.S.A. = 2 [x² + x²] = 4x²
3. Volume = (Area of base) × Height = (x²) · x = x³
4. Length of diagonal = \( x\sqrt{3} \)

(c) Cylinder

Curved surface area (C.S.A.): It is the area of the surface from which the cylinder is formed. When we cut this cylinder, we get a rectangle with length 2πr and height h units.

1. C.S.A. of cylinder = (2πr) × h = 2πrh.
2. Total Surface Area (T.S.A.) = C.S.A. + circular top and bottom = 2πrh + 2πr² = 2πr (h + r) sq. units.
3. Volume of cylinder = Area of base × height = (πr²) × h = πr²h cubic units.

(d) Cone

1. C.S.A. = πrℓ
2. T.S.A. = C.S.A. + area of base = πrℓ + πr² = πr (ℓ + r)
3. Volume = \( \dfrac{1}{3}\pi r^2 h \)

Where h = height, r = radius of base and ℓ = slant height, with ℓ² = r² + h².

(e) Sphere

T.S.A. = S.A. = 4πr²
Volume = \( \dfrac{4}{3}\pi r^3 \)

(f) Hemisphere

1. C.S.A. = 2πr²
2. T.S.A. = C.S.A. + area of circular top = 2πr² + πr² = 3πr²
3. Volume = \( \dfrac{2}{3}\pi r^3 \)

Solved Examples

Example 2. The height of a cylinder is 15 cm. Its curved surface area is 660 sq. cm. Find the diameter.
Answer:
Curved surface area of a cylinder = 2πrh = 660 cm².
2 × \( \dfrac{22}{7} \) × r × 15 = 660, so r = \( \dfrac{660 \times 7}{2 \times 22 \times 15} \) = 7 cm.
Therefore, diameter of the cylinder = 14 cm

Example 4. The sides of a cube are increased by 100%. Find the percent increase in the volume of the cube.
Answer:
Side = a, so volume V = a³.
When the side is increased by 100%, the new side becomes 2a and the new volume V′ = (2a)³ = 8a³ = 8V.
% increase in volume = \( \dfrac{8V - V}{V} \times 100\% \) = 700%

Example 6. How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m?
Answer:
Slant height of the conical tent, ℓ = \( \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{625} \) = 25 m.
Curved surface area of the conical tent = πrℓ = \( \dfrac{22}{7} \) × 7 × 25 = 550 m².
Required length of 5 m wide cloth = \( \dfrac{\text{Area}}{\text{Width}} = \dfrac{550}{5} \) = 110 m.

Example 7. If the total surface area of a solid hemisphere is 462 cm², find its volume. (Take π = \( \dfrac{22}{7} \))
Answer:
Total surface area of the hemisphere = 3πr² = 462 cm².
r² = \( \dfrac{462 \times 7}{3 \times 22} \) = 7 × 7, so r = 7 cm.
Hence, the volume of the hemisphere = \( \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 7 \) cm³ = 718.67 cm³ (approx).

Example 8. The total surface area of a cube is numerically equal to the surface area of a sphere. Find the ratio of their volumes (cube : sphere).
Answer:
T.S.A. of a cube = 6a² and S.A. of a sphere = 4πr². Equating, 6a² = 4πr², so a² = \( \dfrac{2\pi}{3} r^2 \) and a = \( \sqrt{\dfrac{2\pi}{3}}\, r \).
Volume of the cube = a³ = \( r^3 \left( \dfrac{2\pi}{3} \right)^{3/2} \) and volume of the sphere = \( \dfrac{4}{3}\pi r^3 \).
Ratio = \( \dfrac{(2\pi/3)^{3/2}}{4\pi/3} = \dfrac{1}{2}\sqrt{\dfrac{2\pi}{3}} = \) \( \sqrt{\dfrac{\pi}{6}} \).

Example 9. How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm?
Answer:
Volume of the spherical ball of radius 8 cm = \( \dfrac{4}{3}\pi \times 8^3 \) cm³ and volume of each smaller ball of radius 1 cm = \( \dfrac{4}{3}\pi \times 1^3 \) cm³.
Let n be the number of smaller balls. The volume of the larger ball is equal to the sum of the volumes of the n smaller balls.
\( \dfrac{4}{3}\pi \times n = \dfrac{4}{3}\pi \times 8^3 \), so n = 8³ = 512. Hence, the required number of balls = 512.

Check Your Progress

10. From the four corners of a rectangular sheet of paper 4 squares of side 5 cm are cut off. The length and breadth of the paper are 35 cm, 25 cm. The remaining part is folded in such a way to make a cuboid. Find the volume of the cuboid.

11. A right angled triangle has the sides containing the right angle 6 cm and 8 cm. This triangle is revolved about its hypotenuse. Find the volume of the solid generated by the revolution. (Take π = 22/7)

12. The surface areas of 3 faces of a cuboid are 36 sq. cm, 49 sq. cm and 64 sq. cm. Find the volume of the cuboid.

13. A cone has volume 297 cubic cm. The height is 14 cm. Find the diameter of the base.

14. The curved surface area of a cone is 550 sq. cm. Diameter is 14 cm. Find the height.

15. There are two cones. The curved surface area of the first one is twice that of the second. The slant height of the second one is twice that of the first. Find the ratio of their radii.

16. The radius of a sphere is increased 4 times. Find the increase in the volume.

17. Two cylinders have radii in the ratio 1 : 2 and heights in the ratio 2 : 1. Find the ratio of their volumes.

18. Find the length of the longest stick that can be kept in a room of 3 m by 4 m by 5 m.

Answers
1. 1875 cubic cm
2. 241.37 cubic cm
3. 336 cubic cm
4. 9 cm
5. 24 cm
6. 4 : 1
7. The volume becomes 64 times
8. 1 : 2
9. \( 5\sqrt{2} \) m