# Class 8 RD Sharma – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) – Exercise 21.3 | Set 1

**Question 1. Find the surface area of a cuboid whose**

**(i) length = 10 cm, breadth = 12 cm, height = 14 cm****(ii) length = 6 dm, breadth = 8 dm, height = 10 dm****(iii) length = 2 m, breadth = 4 m, height = 5 m****(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.**

**Solution :**

i)length = 10 cm, breadth = 12 cm, height = 14 cm

Given, the Length of cuboid = 10 cm

The Breadth of cuboid = 12 cm

The Height of cuboid = 14 cm

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(10 Ã— 12 + 12 Ã— 14 + 14 Ã— 10) = 856 cm^{2}

Hence, the surface area of given cuboid is856 cm^{2}

ii)length = 6 dm, breadth = 8 dm, height = 10 dm

Given, the Length of cuboid = 6 dm

The Breadth of cuboid = 8 dm

The Height of cuboid = 10 dm

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(6 Ã— 8 + 8 Ã— 10 + 10 Ã— 6) = 376 dm^{2}

Hence, the surface area of given cuboid is 376 dm^{2}

iii)length = 2 m, breadth = 4 m, height = 5 m

Given, the Length of cuboid = 2 m

The Breadth of cuboid = 4 m

The Height of cuboid = 5 m

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(2 Ã— 4 + 4 Ã— 5 + 5 Ã— 2) = 76 m^{2}

Hence, the surface area of given cuboid is 76m^{2}

iv)length = 3.2 m, breadth = 30 dm, height = 250 cm

Given, the Length of cuboid = 3.2 m = 32 dm

The Breadth of cuboid = 30 dm

The Height of cuboid = 250 dm = 25 dm

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(32 Ã— 30 + 30 Ã— 25 + 25 Ã— 32) = 5020 dm^{2}

Hence, the surface area of given cuboid is 5020dm^{2}

**Question 2. Find the surface area of a cube whose edge is**

**(i) 1.2 m****(ii) 27 cm****(iii) 3 cm****(iv) 6 m****(v) 2.1 m**

**Solution:**

i)1.2 m

Given, the Edge of cube = 1.2 m

And the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(1.2)^{2}= 8.64 m^{2}

Hence, the surface area of a given cube is 8.64m^{2}

ii)27 cm

Given, the Edge of cube = 27 cm

And the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(27)^{2}= 4374 cm^{2}

Hence, the surface area of a given cube is 4374cm^{2}

iii)3 cm

Given, the Edge of cube = 3 cm

And the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(3)^{2}= 54 cm^{2}

Hence, the surface area of a given cube is 54cm^{2}

iv)6 m

Given, the Edge of cube = 6 m

And the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(6)^{2}= 216 m^{2}

Hence, the surface area of a given cube is 216m^{2}

v)2.1 m

Given, the Edge of cube = 2.1 m

And the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(2.1)^{2}= 26.46 m^{2}

Hence, the surface area of a given cube is 26.46m^{2}

**Question 3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area?**

**Solution :**

Given, the Length of cuboidal box = 5 cm

The Breadth of cuboidal box = 5 cm

The Height ofthecuboidal box = 4 cm

And, Surface area of cuboidal box =2(l Ã— b + b Ã— h + h Ã— l)

= 2(5 Ã— 5 + 5 Ã— 4 + 4 Ã— 5) = 130 cm^{2}

Hence, the surface area ofthegiven cuboidal box is 130cm^{2}

**Question 4. Find the surface area of a cube whose volume is**

**(i) 343 m ^{3}**

**(ii) 216 dm**

^{3}**Solution:**

i)Given, the volume of cube = 343 m^{3}

It means (side)^{3}= 343

So, side = 7 m

And, the Surface area of cube =6 Ã— side^{2}

= 6Ã—(7)^{2}= 294 m^{2}

Hence, the surface area of a given cube is 294m^{2}

ii)Given, the volume of cube = 216 dm^{3}

It means (side)^{3}= 216

So, side = 6 dm

And, the Surface area of cube =6 Ã— side^{2}

= 6Ã—(6)^{2}= 216 dm^{2}

Hence, the surface area of a given cube is 216dm^{2}

**Question 5. Find the volume of a cube whose surface area is**

**(i) 96 cm ^{2}**

**(ii) 150 m**

^{2}**Solution:**

i)Given, the surface area of cube = 96 cm^{2}

It means6 Ã— side^{2}= 96

So, side = 4 cm

And, the Volume of cube = (side)^{3}

= (4)^{3}= 64 cm^{3}

Hence, the volume of the given cube is 64 cm^{3}

ii)Given, the surface area of cube = 150 m^{2}

It means6 Ã— side^{2}= 150

So, side = 5 m

And, the Volume of cube = (side)^{3}

= (5)^{3}= 125 m^{3}

Hence, the volume of the given cube is 125 m^{3}

**Question 6. The dimensions of a cuboid are in the ratio 5 : 3 : 1 and its total surface area is 414 m**^{2}. Find the dimensions?

^{2}. Find the dimensions?

**Solution:**

Given, the dimensions of a cuboid are in the ratio 5 : 3 : 1

The total surface area is 414 m^{2}

So, the Length of cuboid = 5a

The Breadth of cuboid = 3a

The Height of cuboid = 1a

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

414 = 2(5a Ã— 3a + 3a Ã— a + a Ã— 5a)

414 = 46a^{2}

So, a = 3 m

We can conclude, Length = 5a = 15 m

The Breadth = 3a = 9 m

The Height = a = 3 m

Hence, the dimensions of a cuboid are 15 m, 9 m, and 3 m.

**Question**** 7. ****Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m**,** and height 15 cm?**

**Solution:**

Given, the Length of Cuboidal cardboard = 25 cm

The Breadth of Cuboidal cardboard = 0.5 m = 50 cm

The Height of Cuboidal cardboard = 15 cm

Area of cardboard needed = Area of Cuboid

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(25 Ã— 50 + 50 Ã— 15 + 15 Ã— 25) = 4750 cm^{2}

Hence, the area of cardboard required is 4750 cm^{2}

**Question 8. Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm?**

**Solution :**

Given, the Edge of cube = 12 cm

The surface area of a wooden box = Surface area of Cube

And, the Surface area of cube =6 Ã— edge^{2}

= 6Ã—(12)^{2}= 864 cm^{2}

Hence, the surface area of a wooden box is 864 cm^{2}

**Question 9. The dimensions of an oil tin are 26 cmÃ— 26 cmÃ— 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square meter of the tin sheet costs Rs. 10, find the cost of the tin sheet used for these 20 tins?**

**Solution :**

Given, the Length of oil tin = 26 cm

The Breadth of oil tin = 26 cm

The Height of oil tin = 45 cm

Area of tin sheet required to make 1 oil tin = Surface area of Cuboid

And, Surface area of cuboid =2(l Ã— b + b Ã— h + h Ã— l)

= 2(26 Ã— 26 + 26 Ã— 45 + 45 Ã— 26) = 6032 cm^{2}

And, Area of tin sheet required to make 20 oil tins = 20 Ã— 6032 = 120640 cm^{2}

= 12.064 m^{2}

Since, 1 m^{2}Tin cost = Rs 10

So, 12.064 m^{2}Tin cost = Rs 10 Ã— 12.064

= Rs 120.64

Hence, the area of tin required to make 20 oil tin is 12.064 m^{2}and the cost of making 20 oil tin is Rs 120.64

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