RD Sharma Class 8 Solutions Chapter 21 Mensuration Ex 21.3



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TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 21
Chapter NameMensuration
ExerciseEx 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) cm2
= 2(120+ 168 + 140) cm2
= 2 x 428 = 856 cm2
(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) dm2
= 2(48 + 80 + 60) dm2 = 2 x 188 = 376 dm2
(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) dm2
= 2(960 + 750 + 800) dm2
= 2 x 2510 = 5020 dm2

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 = 6a2= 6 x (1,2)2 m2
= 6 x 1.44 = 8.64 m2
(ii) Edge of cube (a) = 27 cm
∴ Surface area = 6a2 = 6 x (27)2 m2
= 6 x 729 = 4374 m2
(iii) Edge of cube (a) = 3 cm
Surface area = 6a2 = 6 x (3)2 m2
= 6×9 cm2 = 54 cm2
(iv) Edge of cube (a) = 6 m
∴ Surface area = 6a2 = 6 x (6)2 m2
= 6 x 6 x 6 = 216 m2
(v) Edge of the cube (a) = 2.1 m
∴ Surface area = 6a2 = 6 x (2.1)2 m2
= 6 x 4.41 = 26.46 m2

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) cm2
= 2 (25 + 20 + 20)
= 2 x 65 cm2
= 130 cm2

Question 4.
Find the surface area of a cube whose volume is :
(i) 343 m3
(ii) 216 dm3.
Solution:
(i) Volume of a cube = 343 m3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 1

Question 5.
Find the volume of a cube whose surface area is
(i) 96 cm2
(ii) 150 m2.
Solution:
(i) Surface area of a cube = 96 cm2
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 2

Question 6.
The dimensions of a cuboid are in the ratio 5:3:1 and its total surface area is 414 m2. 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
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 3

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) cm2
= 2(1250+ 750+ 375) cm2
= 2(2375)
= 4750 cm2

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 = 6a2 = 6(12)2 cm2
= 6 x 144 = 864 cm2

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) m2
= 2(0.0676 + 0.117 + 0.117) m2
= 2(0.3016) = 0.6032 m2
Sheet required for such 20 tins
= 0.6032 x 20= 12.064 m2
Cost of 1 m2 tin sheet = 10 m
∴ Total cost = Rs 12.064 x 10 = Rs 120.64
and area of sheet = 12.064 m2 = 120640 cm2

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 m2 = 2 x 19×5 = 190 m2
∴ Total area = 88 m2 + 190 m2 = 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 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 m2
and area of its walls = 2(l + b) x h
= 2(20 + 15) x 3 = 2 x 35 x 3 m2 = 210 m2
∴ Total area = 300 + 210 = 510 m2
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 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 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)
= l2b2h2 = (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 m2 = 2 x 7.5 x 3.5 m2 = 52.5 m2
Area of ceiling = l x b = 4.5 x 3 = 13.5 m2
∴ Total area = 52.5 + 13.5 m2 = 66 m2
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 m2 and lateral surface area its 30 m2. Find the area of its base.
Solution:
Total surface area of cuboid = 50 m2
Lateral surface area = 30 m2
∴ Area of floor and ceiling = 50 – 30 = 20 m2
But area of floor = area of ceiling
∴ Area of base (floor) = 202 = 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 white-washing the walls at the rate of Rs 1.50 per m2.
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 m2 = 2 x 13 x 3.5 = 91 m2
Area of doors and windows = 17 m2
∴ Remaining area of walls = 91 – 17 = 74 m2
Rate of whitewashing = Rs 1.50 per m2
∴ 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 m2 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 m2
= 45 m2
A size of each windows = 1.5 m x 1 m
∴ Area of 10 windows = 1.5 m x 1 x 10= 15 m2
Total cost of whitewashing the walls = Rs 2385.60
Rate of whitewashing = Rs 1.20 per m2
∴ Area of walls which are whitewashed
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 4

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