In this chapter, we provide RD Sharma Class 8 Solutions Chapter 21 Mensuration Ex 21.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 8 Solutions Chapter 21 Mensuration Ex 21.3 pdf, free RD Sharma Class 8 Solutions Chapter 21 Mensuration Ex 21.3 book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 8 |

Subject | Maths |

Chapter | Chapter 21 |

Chapter Name | Mensuration |

Exercise | Ex 21.3 |

**RD Sharma Solutions for Class 8 Chapter 21 Mensuration Ex 21.3** **Download PDF**

**Question 1.****Find the surface area of a cuboid whose :****(i) length = 10 cm, breadth = 12 cm and height = 14 cm****(ii) length = 6 dm, breadth = 8 dm, height = 10 dm****(iii) length = 2 m, breadth = 4 m and height = 5 m****(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.****Solution:****(i)** Length of cuboid (l) = 10 cm

Breadth (b) = 12 cm

Height (h) = 14 cm

∴ Surface area = 2(1 × b + b × h + h × l)

= 2(10 x 12 + 12 x 14 + 14 x 10) cm^{2}

= 2(120+ 168 + 140) cm^{2}

= 2 x 428 = 856 cm^{2}**(ii)** Length of cuboid (l) = 6 dm

Breadth (b) = 8 dm

Height (h) = 10 dm

∴ Surface area = 2 ( l × b + b x h + h× l)

= 2(6 x 8 + 8 x 10 + 10 x 6) dm^{2}

= 2(48 + 80 + 60) dm^{2} = 2 x 188 = 376 dm^{2}**(iii)** Length of cuboid (l) = 2 m

Breadth (b) = 4 m

Height (h) = 5 m

∴ Surface area = 2(l × b + b × h + h × l)

= 2(2 x 4 + 4 x 5 + 5 x 2) m2

= 2(8 + 20 + 10) m2 = 76 m2**(iv)** Length of cuboid (l) = 3.2 m = 32 dm

Breadth (b) = 30 dm

Height (h) = 250 cm = 25 dm

∴ Surface area = 2(1 x b + b x h + h x l)

= 2(32 x 30 + 30 x 25 + 25 x 32) dm^{2}

= 2(960 + 750 + 800) dm^{2}

= 2 x 2510 = 5020 dm^{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) 2.1m****Solution:****(i)** Edge of the cube (a) = 1.2 m

∴ Surface area = 6a^{2}= 6 x (1,2)^{2} m^{2}

= 6 x 1.44 = 8.64 m^{2}**(ii)** Edge of cube (a) = 27 cm

∴ Surface area = 6a^{2} = 6 x (27)^{2} m^{2}

= 6 x 729 = 4374 m^{2}**(iii)** Edge of cube (a) = 3 cm

Surface area = 6a^{2} = 6 x (3)^{2} m^{2}

= 6×9 cm^{2} = 54 cm^{2}**(iv)** Edge of cube (a) = 6 m

∴ Surface area = 6a^{2} = 6 x (6)2 m^{2}

= 6 x 6 x 6 = 216 m^{2}**(v)** Edge of the cube (a) = 2.1 m

∴ Surface area = 6a2 = 6 x (2.1)2 m^{2}

= 6 x 4.41 = 26.46 m^{2}

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

Length of cuboid box (l) = 5 cm

Breadth (b) = 5 cm

and height (h) = 4 cm

∴ Surface area = 2 (l x b + b x h + h x l)

= 2 (5 x 5 + 5 x 4 + 4 x 5) cm^{2}

= 2 (25 + 20 + 20)

= 2 x 65 cm^{2}

= 130 cm^{2}

**Question 4.****Find the surface area of a cube whose volume is :****(i) 343 m ^{3}**

**(ii) 216 dm**.

^{3}**Solution:**

(i) Volume of a cube = 343 m

^{3}

**Question 5.****Find the volume of a cube whose surface area is****(i) 96 cm ^{2}**

**(ii) 150 m**.

^{2}**Solution:**

(i) Surface area of a cube = 96 cm

^{2}

**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.**

**Solution:**

Ratio in .dimensions = 5 : 3 : 1

Let length (l) = 5x

breadth (b) = 3x

and height (h) = x

∴ Surface area = 2(1 x b + b x h + h x l)

= 2(5x x 3x + 3x x x + x x 5x)

= 2(15×2 + 3×2 + 5×2) = 2 x 23×2 = 46×2

**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:**

Length of cardboard (l) = 25 cm

Breadth (b) = 0.5 m = 50 cm

Height (h)= 15 cm.

∴ Surface area of cardboard = 2 (l x b + b x h + h x l)

= 2(25 x 50 + 50 x 15 + 15 x 25) cm^{2}

= 2(1250+ 750+ 375) cm^{2}

= 2(2375)

= 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:**

Edge of cubic wooden box = 12 cm

∴ Surface area = 6a^{2} = 6(12)^{2} cm^{2}

= 6 x 144 = 864 cm^{2}

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

Length of tin (l) = 26 cm = 0.26 m

Breadth (b) = 26 cm = 0.26 m

Height (h) = 45 cm = 0.45 m

∴ Surface area = 2(l x b + b x h +h xl)

= 2(0.26 x 0.26 + 0.26 x 0.45 + 0.45 x 0.26) m^{2}

= 2(0.0676 + 0.117 + 0.117) m^{2}

= 2(0.3016) = 0.6032 m^{2}

Sheet required for such 20 tins

= 0.6032 x 20= 12.064 m^{2}

Cost of 1 m^{2} tin sheet = 10 m

∴ Total cost = Rs 12.064 x 10 = Rs 120.64

and area of sheet = 12.064 m^{2} = 120640 cm^{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:**

Length of room (l) = 11 m

Width (b) = 8 m

and height (h) = 5 m

Area of floor = l x b = 11 x8 = 88m2

Area of four walls = 2 (l + b) x h

= 2(11 + 8) x 5 m^{2} = 2 x 19×5 = 190 m^{2}

∴ Total area = 88 m^{2} + 190 m^{2} = 278 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 metre.****Solution:**

Length of pool (l) = 20 m

Breadth (b) = 15 m

and Depth (h) = 3 m.

Area of floor = l x b = 20 x 15 = 300 m^{2}

and area of its walls = 2(l + b) x h

= 2(20 + 15) x 3 = 2 x 35 x 3 m^{2} = 210 m^{2}

∴ Total area = 300 + 210 = 510 m^{2}

Rate of repairing it = Rs 25 per sq. metre

∴ Total cost = Rs 25 x 510 = 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 four walls of the room.****Solution:**

Perimeter of floor = 30 m

i.e. 2(1 + b) = 30 m

Height = 3 m

∴ Area of four walls = Perimeter x height = 30 x 3 = 90 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:**

Let length of the room = l

and breadth = b

and height = h

Volume = l x b x h

Area of floor = l x b = lb.

Area of two adjacent walls = hl x bh.

∴ Product of areas of floor and two adjacent walls of the room = lb (hi x bh)

= l^{2}b^{2}h^{2} = (l.b.h)^{2} = (Volume)^{2}

Hence proved

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

Length of room (l) = 4.5 m

Width (b) = 3 m

and height (h) = 350 cm = 3.5 m

∴ Area of walls = 2(l + b) x h

= 2(4.5 + 3) x 3.5 m^{2} = 2 x 7.5 x 3.5 m^{2} = 52.5 m^{2}

Area of ceiling = l x b = 4.5 x 3 = 13.5 m^{2}

∴ Total area = 52.5 + 13.5 m^{2} = 66 m^{2}

Rate of plastering = Rs 8 per sq. m

∴ Total cost = Rs 8 x 66 = Rs 528

**Question 15.****A cuboid has total surface area of 50 m ^{2} and lateral surface area its 30 m^{2}. Find the area of its base.**

**Solution:**

Total surface area of cuboid = 50 m

^{2}

Lateral surface area = 30 m

^{2}

∴ Area of floor and ceiling = 50 – 30 = 20 m

^{2}

But area of floor = area of ceiling

∴ Area of base (floor) = 202 = 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 white-washing the walls at the rate of Rs 1.50 per m^{2}.**

**Solution:**

Length of room (l) = 7 m

Breadth (b) = 6 m

and height (h) = 3.5 m

∴ Area of four walls = 2(1 + b) x h

= 2(7 + 6) x 3.5 m

^{2}= 2 x 13 x 3.5 = 91 m

^{2}

Area of doors and windows = 17 m2

∴ Remaining area of walls = 91 – 17 = 74 m

^{2}

Rate of whitewashing = Rs 1.50 per m

^{2}

∴ Total cost = 74 x Rs 1.50 = 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 3 m x 1.5 m and 10 windows each of size 1.5 m x l m. If the cost of the white-washing 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.**

**Solution:**

Length of hall (l) = 80 m

Height (h) = 8 m

Size of each door = 3 m x 1.5 m

∴ Area of 10 doors = 3 x 1,5 x 10 m

^{2}

= 45 m

^{2}

A size of each windows = 1.5 m x 1 m

∴ Area of 10 windows = 1.5 m x 1 x 10= 15 m

^{2}

Total cost of whitewashing the walls = Rs 2385.60

Rate of whitewashing = Rs 1.20 per m

^{2}

∴ Area of walls which are whitewashed

**All Chapter RD Sharma Solutions For Class 8 Maths**

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**All Subject NCERT Solutions For Class 8**

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