# Class 8 RD Sharma – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) – Exercise 21.3 | Set 2

**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:**

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The Breadth of classroom = 8 m

The Height of the classroom = 5 m

Area of floor =length × breadth = 11 × 8 = 88 m^{2}

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

= 2 (11 × 5 + 8 × 5)

= 190 m^{2}

The total area of four walls and floor = Area of floor +Area offour walls (including doors, windows)= 88 + 190 = 278 m^{2}

Hence, thesum of the areas of its floor and the four walls is278 m^{2}

**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 m^{2}

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

= 2 (20 × 3 + 15 × 3)

= 210 m^{2}

The total area of swimming pool = Area of floor +Area offour walls= 300 + 210 = 510 m^{2}

Since, the cost of repairing 1 m^{2}swimming pool area = Rs 25

So, the cost of repairing 1 m^{2}swimming pool area = Rs 25 × 510 =Rs 12750

Hence, thecost of repairing the floor and wall of the swimming pool isRs 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 m^{22}

Hence, the area of the four walls of the room is90 m^{2}

**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:**

Letlbe the length of the cuboid

Letbbe the breadth of cuboid

Lethbe 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) = lbh^{2}

And, finally the product of the areas of the floor and two adjacent walls of a cuboid = lb Xlbh^{2 }= (l × b × h)^{2}

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

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

= 2 (4.5 × 3.5 + 3 × 3.5)

= 52.5 m^{2}

So, the Sum of the area of ceiling and area of four walls = 13.5 + 52.5 =66m^{2}

Also, the Cost of plastering room 1m^{2}area = Rs 8

So, the Cost of plastering room 66m^{2}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 m**^{2} and a lateral surface area is 30 m^{2}. Find the area of its base?

^{2}and a lateral surface area is 30 m

^{2}. Find the area of its base?

**Solution:**

Given, the Total surface area of cuboid = 50 m^{2}

The Lateral surface area of cuboid = 30 m^{2}

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 m^{2}

Hence, the area of the base of cuboid =10 m^{2}

**Question 16. A classroom is 7 m long, 6 m broad, and 3.5 m high. Doors and windows occupy an area of 17 m**^{2}. What is the cost of whitewashing the walls at the rate of Rs 1.50 per m^{2}?

^{2}. What is the cost of whitewashing the walls at the rate of Rs 1.50 per m

^{2}?

**Solution:**

Given, the Length of classroom = 7 m

The Breadth of classroom = 6 m

The height oftheclassroom = 3.5 mThe area

occupied by doors and windows= 17 m^{2}

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

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

So,thearea of 4 walls excludingdoors & windows = 91 – 17 = 74 m^{2}

Also, the cost of whitewashing 1m^{2}classroom = Rs 1.50

So, the cost of whitewashing 74m^{2}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 m**^{2} is Rs 2385.60, find the breadth of the hall?

^{2}is Rs 2385.60, find the breadth of the hall?

**Solution:**

Given, the Length of central hall = 80 m

The height ofthecentral hall = 8 m

And letthebbe the Breadth ofhall

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 m^{2}

And, the Area of 10 doors = 4.5 × 10 = 45 m^{2}

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 m^{2}

And, the Area of 10 windows = 1.5 × 10 = 15 m^{2}

So, the total area occupied by doors & windows = 45 + 15 = 60 m^{2}

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

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

So, area of 4 walls excludingdoors & windows = 2(640+8b) – 60 = (1220 + 16b) m^{2}

Also, the cost of whitewashing 1m^{2}central hall = Rs 1.20

So, the cost of whitewashing(1220 + 16b)m^{2}central hall =(1220 + 16b)×1.20

Total whitewashing cost = Rs2385.60

Itmeans,

2385.60 = (1220 + 16b)×1.20

b = 48 m

Hence, the breadth of central hall is 48 m