ICSE Class 9 Maths Chapter 21 Volume and Surface Area of Solids

Volume and Surface Area of Solids

Points To Remember

1. Solids. The bodies occupying space are called solids. The solid bodies occur in various shapes, such as a cuboid, a cube, a cylinder, a cone and a sphere, etc.

2. Volume of a Solid. The space occupied by a solid body is called its volume. The units of volume are cubic cm (i.e., cm³) or cubic metres (i.e., m³), etc.

3. Cuboid. A rectangular solid bounded by six rectangular plane faces is called a cuboid. A cuboid has 6 rectangular faces, 12 edges and 8 vertices.

4. Cube. A cuboid whose length, breadth and height are all equal is called a cube. Each edge of a cube is called its side. It has 6 square faces, 12 edges and 8 vertices.

5. Formulae

1. Cuboid. Let length = l units, breadth = b units and height = h units. Then,

(i) Volume of the cuboid = (l × b × h) cubic units.

(ii) Diagonal of the cuboid = \(\sqrt{l^2 + b^2 + h^2}\) units.

(iii) Total Surface Area of the cuboid = 2(lb + bh + lh) sq. units.

(iv) Lateral Surface Area of the cuboid = [2(l + b) × h] sq. units.

(v) Area of 4 walls of a room = [2(l + b) × h] sq. units.

2. Cube. Let edge of a cube = a units. Then,

(i) Volume of the cube = a³ cubic units.

(ii) Diagonal of the cube = (a\(\sqrt{3}\)) units.

(iii) Total Surface Area of the cube = (6a²) sq. units.

(iv) Lateral Surface Area of the cube = (4a²) sq. units.

Exercise 21 (A)

Q. 1. Find the volume; total surface area and the lateral surface area of a rectangular solid having:

(i) length = 8.5 m, breadth = 6.4 m and height = 50 cm.

(ii) length = 5.6 dm, breadth = 22.5 dm and height = 1 m.

Sol. (i) Length = 8.5 m, Breadth = 6.4 m and Height = 50 cm = 0.5 m

(a) Volume = Length × Breadth × Height = 8.5 m × 6.4 m × 0.5 m = 27.2 m³

(b) Total surface area = 2(lb + bh + lh) = 2[8.5 × 6.4 + 6.4 × 0.5 + 0.5 × 8.5] m² = 2[54.4 + 3.2 + 4.25] m² = 2 × 61.85 = 123.7 m²

(c) Lateral surface area = 2(l + b)h = 2(8.5 + 6.4) × 0.5 m² = 2 × 14.9 × 0.5 = 14.9 m² Ans.

(ii) Length (l) = 5.6 dm, Breadth (b) = 2.5 dm, Height (h) = 1 m = 10 dm

(a) Volume = lbh = 5.6 × 2.5 × 10 = 140 dm³

(b) Total surface area = 2(lb + bh + lh) = 2[5.6 × 2.5 + 2.5 × 10 + 10 × 5.6] dm² = 2[14.0 + 25 + 56] dm² = 2 × 95 = 190 dm²

(c) Lateral surface area = 2(l + b) × h = 2(5.6 + 2.5) × 10 dm² = 2 × 8.1 × 10 = 162 dm² Ans.

Q. 2. The volume of a rectangular wall is 33 m³. If its length is 16.5 m and height 8 m, find the width of the wall.

Sol. Volume of rectangular wall = 33 m³, Length of wall (l) = 16.5 m, Height of wall (h) = 8 m, Let width of wall = b, then lbh = volume

16.5 × 8 × b = 33

b = \(\frac{33}{16.5 \times 8}\) = \(\frac{1}{4}\) m = 0.25 m Ans.

Q. 3. Find the number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm, required to construct a wall 6 m long, 5 m high and 50 cm thick, while the cement and the sand mixture occupies \(\frac{1}{20}\)th of the volume of the wall.

Sol. Volume of one brick = 25 cm × 12.5 cm × 7.5 cm = \(\frac{25}{100}\) × \(\frac{12.5}{100}\) × \(\frac{7.5}{100}\) m³ = \(\frac{1}{4}\) × \(\frac{1}{8}\) × \(\frac{3}{40}\) m³ = \(\frac{3}{1280}\) m³

Length of wall (l) = 6 m, Height of wall (h) = 5 m, and Thickness (b) = \(\frac{50}{100}\) m = \(\frac{1}{2}\) m

Volume = lbh = 6 × \(\frac{1}{2}\) × 5 = 15 m³

Volume of cement and sand = \(\frac{1}{20}\) of 15 m³ = \(\frac{3}{4}\) m³

Volume of bricks = 15 - \(\frac{3}{4}\) = \(\frac{60 - 3}{4}\) = \(\frac{57}{4}\) m³

No. of bricks = \(\frac{\text{Volume of total bricks}}{\text{Volume of one brick}}\) = \(\frac{\frac{57}{4}}{\frac{3}{1280}}\) = \(\frac{57}{4}\) × \(\frac{1280}{3}\) = 19 × 320 = 6080 Ans.

Q. 4. A class room is 12.5 m long, 6.4 m broad and 5 m high. How many students can accommodate if each student needs 1.6 m² of floor area? How many cubic metres of air would each student get?

Sol. Length of room (l) = 12.5 m, Width of room (b) = 6.4 m, and Height (h) = 5 m

Volume of air inside the room = lbh = 12.5 × 6.4 × 5 m³ = 400 m³

Area of floor of the room = l × b = 12.5 × 6.4 m² = 80 m²

For each student area required = 1.6 m²

No. of students = \(\frac{80}{1.6}\) = \(\frac{80 \times 10}{16}\) = 50

and each student required the air = \(\frac{\text{Volume of air}}{\text{No. of students}}\) = \(\frac{400}{50}\) = 8 m³ Ans.

Q. 5. Find the length of the longest rod that can be placed in a room measuring 12 m × 9 m × 8 m.

Sol. Length of room (l) = 12 m, Breadth (b) = 9 m, and height (h) = 8 m

The longest rod required to place in the room = \(\sqrt{l^2 + b^2 + h^2}\) = \(\sqrt{(12)^2 + (9)^2 + (8)^2}\) = \(\sqrt{144 + 81 + 64}\) m = \(\sqrt{289}\) = 17 m Ans.

Q. 6. The volume of a cuboid is 14400 cm³ and its height is 15 cm. The cross-section of the cuboid is a rectangle having its sides in the ratio 5:3. Find the perimeter of the cross-section.

Sol. Volume of cuboid = 14400 cm³, Height (h) = 15 cm

Length × Breadth = \(\frac{\text{Volume}}{h}\) = \(\frac{14400}{15}\) cm² = 960 cm²

Ratio in remaining sides = 5:3

Let length = 5x and breadth = 3x

5x × 3x = 960

15x² = 960

x² = 64 = (8)²

x = 8

Length = 5x = 8 × 5 = 40 cm and breadth = 3x = 8 × 3 = 24 cm

Perimeter of rectangular cross-section = 2(l + b) = 2(40 + 24) cm = 2 × 64 = 128 cm Ans.

Q. 7. The area of path is 6500 m². Find the cost of covering it with gravel 14 cm deep at the rate of Rs. 5.60 per cubic metre.

Sol. Area of path = 6500 m²

Depth of gravel = 14 cm = \(\frac{14}{100}\) m

Volume of gravel = Area × Depth = 6500 × \(\frac{14}{100}\) = 910 m³

Rate of covering the gravel = Rs. 5.60 per m³

Total cost = Rs. 10 × 5.60 = \(\frac{910 \times 560}{100}\) = Rs. 5096 Ans.

Q. 8. The cost of papering the four walls of a room 12 m long at Rs. 6.50 per square metre is Rs. 1638 and the cost of matting the floor at Rs. 3.50 per square metre is Rs. 378. Find the height of the room.

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