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Class 8 RD Sharma – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) – Exercise 21.3 | Set 2
  • Last Updated : 28 Dec, 2020

Question 10. A classroom is 11 m long, 8 m wide, and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows, etc.)?

Solution:

Given, the Length of classroom = 11 m

The Breadth of classroom = 8 m

The Height of the classroom = 5 m

Area of floor = length × breadth = 11 × 8 = 88 m2

And, the Area of four walls (including doors, windows) = 2 (lh + bh)

= 2 (11 × 5 + 8 × 5)

= 190 m2



The total area of four walls and floor = Area of floor + Area of four walls (including doors, windows) = 88 + 190 = 278 m2

Hence, the sum of the areas of its floor and the four walls is 278 m2

Question 11. A swimming pool is 20 m long 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs. 25 per square meter?

Solution:

Given, the Length of the swimming pool = 20 m

The Breadth of the swimming pool = 15 m

The Height of the swimming pool = 3 m

Area of the swimming pool floor = length × breadth = 20 × 15 = 300 m2

Area of the swimming pool walls = 2 (lh + bh)

= 2 (20 × 3 + 15 × 3)

= 210 m2

The total area of swimming pool = Area of floor + Area of four walls = 300 + 210 = 510 m2



Since, the cost of repairing 1 m2 swimming pool area = Rs 25

So, the cost of repairing 1 m2 swimming pool area = Rs 25 × 510 = Rs 12750

Hence, the cost of repairing the floor and wall of the swimming pool is Rs 12750

Question 12. The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of the four walls of the room?

Solution:

Given, the perimeter of a room floor = 30m

It means 2(l + b) = 30

So, l + b = 15 m

The height of the room = 3 m

Now, the Area of the room walls = 2 (lh + bh)

= 2h (l + b) = 2 × 3 × 15 = 90 m22

Hence, the area of the four walls of the room is 90 m2

Question 13. Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume?

Solution:

Let l be the length of the cuboid 

Let b be the breadth of cuboid 

Let h be the height of the cuboid 

So, the Area of floor = length X breadth = l × b

And, the product of the area of two adjacent walls of cuboid = (l × h) × (b × h) = lbh2

And, finally the product of the areas of the floor and two adjacent walls of a cuboid = lb X lbh2 = (l × b × h)2

{We know that Volume of cuboid = l × b × h} = (Volume of the cuboid)2

Hence, proved that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume

Question 14. The walls and ceiling of a room are to be plastered. The length, breadth, and height of the room are 4.5 m, 3 m, and 350 cm, respectively. Find the cost of plastering at the rate of Rs. 8 per square meter?

Solution:

Given, the Length of room = 4.5 m

The Breadth of room = 3 m

The Height of room = 350 cm = 3.5 m

Area of the room ceiling = length × breadth = 4.5 × 3 = 13.5 m2

Area of the room walls = 2 (lh + bh)

= 2 (4.5 × 3.5 + 3 × 3.5)

= 52.5 m2

So, the Sum of the area of ceiling and area of four walls = 13.5 + 52.5 = 66m2

Also, the Cost of plastering room 1 m2 area = Rs 8

So, the Cost of plastering room 66 m2 area = Rs 8 × 66 = Rs 528

Hence, the cost of plastering is Rs 528

Question 15. A cuboid has a total surface area of 50 m2 and a lateral surface area is 30 m2. Find the area of its base?

Solution:

Given, the Total surface area of cuboid = 50 m2

The Lateral surface area of cuboid = 30 m2

As we know the lateral surface area of cuboid = Area of 4 walls of cuboid

And, the Total surface area of a cuboid = 2 (Area of Base) + Area of 4 walls 

50 = 2 (Area of Base) + 30

So, Area of Base = 10 m2

Hence, the area of the base of cuboid = 10 m2

Question 16. A classroom is 7 m long, 6 m broad, and 3.5 m high. Doors and windows occupy an area of 17 m2. What is the cost of whitewashing the walls at the rate of Rs 1.50 per m2?

Solution:

Given, the Length of classroom = 7 m

The Breadth of classroom = 6 m

The height of the classroom = 3.5 m

The area occupied by doors and windows = 17 m2

So, area of 4 walls (including doors & windows) = 2(lh + bh) = 2(7 × 3.5 + 6 × 3.5) = 91m2

Since, whitewashing can’t be done on doors and windows,

So, the area of 4 walls excluding doors & windows = 91 – 17 = 74 m2

Also, the cost of whitewashing 1 m2 classroom = Rs 1.50

So, the cost of whitewashing 74 m2 classroom = Rs 1.50 × 74 = Rs 111

Hence, the cost of whitewashing the walls is Rs 111

Question 17. The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3m × 1.5m and 10 windows each of size 1.5m × 1m. If the cost of whitewashing the walls of the hall at the rate of Rs 1.20 per m2 is Rs 2385.60, find the breadth of the hall?

Solution:

Given, the Length of central hall = 80 m

The height of the central hall = 8 m

And let b be the Breadth of the hall

For Doors,

Length of door = 3 m

Width of door = 1.5 m

So, the Area of Door = l × b = 3 × 1.5 = 4.5 m2

And, the Area of 10 doors = 4.5 × 10 = 45 m2

For Windows,

Length of window = 1.5 m

Breadth of window = 1 m

So, the Area of window = l × b = 1.5 × 1 = 1.5 m2

And, the Area of 10 windows = 1.5 × 10 = 15 m2

So, the total area occupied by doors & windows = 45 + 15 = 60 m2

The Area of 4 walls (including doors & windows) = 2(lh + bh) = 2(80 × 8 + b × 8) = 2(640 + 8b) m2

Since, whitewashing can’t be done on doors and windows,

So, area of 4 walls excluding doors & windows = 2(640+8b) – 60 = (1220 + 16b) m2

Also, the cost of whitewashing 1 m2 central hall = Rs 1.20

So, the cost of whitewashing (1220 + 16b) m2 central hall = (1220 + 16b) × 1.20

Total whitewashing cost = Rs 2385.60

It means,

2385.60 = (1220 + 16b) × 1.20

b = 48 m

Hence, the breadth of central hall is 48 m

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