Chapter 10: Surface Areas and Volumes (Exercise 10.1)
Chapter 10: Surface Areas and Volumes (Exercise 10.1)
Question 1:
Find the lateral surface area (LSA) and total surface area (TSA) of the following right prisms:
1. A cube with edge a = 4 cm
2. A cuboid with length l = 8 cm, breadth b = 6 cm, and height h = 5 cm
Solution:
1. For Cube (a = 4 cm):
* Lateral Surface Area (LSA): 4a² = 4 × (4)² = 4 × 16 = 64 cm²
* Total Surface Area (TSA): 6a² = 6 × (4)² = 6 × 16 = 96 cm²
2. For Cuboid (l = 8 cm, b = 6 cm, h = 5 cm):
* Lateral Surface Area (LSA): 2h(l + b) = 2(5)(8 + 6) = 10 × 14 = 140 cm²
* Total Surface Area (TSA): 2(lb + bh + lh) = 2((8×6) + (6×5) + (5×8))
TSA = 2(48 + 30 + 40) = 2(118) = 236 cm²
Question 2:
The total surface area of a cube is 1350 sq. m. Find its volume.
Solution:
* Let the side length of the cube be a.
* Given TSA = 6a² = 1350
a² = 1350 / 6 = 225
a = √225 = 15 m
* Formula for Volume of a cube = a³
Volume = (15)³ = 3375 m³
Question 3:
Find the area of four walls of a room if its length is 12 m, breadth is 10 m, and height is 7.5 m.
Solution:
* The area of four walls of a room equals its Lateral Surface Area (LSA).
Given: l = 12 m, b = 10 m, h = 7.5 m
Area of 4 walls = 2h(l + b)
Area = 2 × 7.5 × (12 + 10)
Area = 15 × 22 = 330 m²
Question 4:
The volume of a cuboid is 1200 cm³. The length is 15 cm and breadth is 10 cm. Find its height.
Solution:
* Formula: Volume of cuboid = length × breadth × height (l × b × h)
Given: Volume = 1200 cm³, l = 15 cm, b = 10 cm
15 × 10 × h = 1200
150 × h = 1200
h = 1200 / 150 = 8 cm
Therefore, the height of the cuboid is 8 cm.
Question 5:
How does the total surface area of a box change if:
1. Each dimension is doubled?
2. Each dimension is tripled?
Solution:
* Original Total Surface Area = S₁ = 2(lb + bh + lh)
1. When dimensions are doubled (2l, 2b, 2h):
New TSA (S₂) = 2[(2l×2b) + (2b×2h) + (2l×2h)]
S₂ = 2[4lb + 4bh + 4lh] = 4 × [2(lb + bh + lh)] = 4 × S₁
The total surface area becomes 4 times the original area.
2. When dimensions are tripled (3l, 3b, 3h):
New TSA (S₃) = 2[(3l×3b) + (3b×3h) + (3l×3h)]
S₃ = 2[9lb + 9bh + 9lh] = 9 × [2(lb + bh + lh)] = 9 × S₁
The total surface area becomes 9 times the original area.