KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3
Question 1.
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Solution:
(a) Length of cuboid (l) = 60 cm
Breadth of cuboid (b) = 40 cm
Height of cuboil (h) = 50 cm
Anount of material required = Total surface area of cuboid
= 2(lb + bh + lh)
= 2(60 × 40 + 40 × 50 + 60 × 50)
= 2[2400 + 2000 + 3000]
= 2 × 7400
= 14800 cm2
Side of cube = 50 cm
(b) Total material required = Total surface area of cube
= 6(side)2
= 6(50)2
= 6 × 2500
= 15000 cm2
Cuboidal boxes required a lesser amount of material to make.
Question 2.
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of suitcase = 80 cm × 48 cm × 24 cm
Surface area of suitcase = 2(lb + bh + lh)
= 2[80 × 48 + 48 × 24 + 80 × 24]
= 2[3840 + 1152 + 1920]
= 2[6912]
= 13824 cm2
Area of cloth required for 100 suitcases = 100 × 13824 = 1382400 cm2
Area of cloth = l × b = l × 96
l × 96 = 1382400
l = 1382400/96 = 14400 cm (approx.)
or length of cloth = 14400/100 m = 144 mts.
Question 3.
Find the side of a cube whose surface area is 600 cm2.
Solution:
Surface area of cube = 600 cm2
⇒ 6(Side)2 = 600
⇒ (Side)2 = 600/6 = 100
⇒ Side = √100 = 10 mts
Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.
Solution:
Measurement of cabinet = 1 m × 2 m × 1.5 m
Surface area of cabinet except base = 2 (lb + bh + lh) – lb
= 2[1 × 2 + 2 × 1.5 + 1.5 × 1] – 1 × 2
= 2[2 + 3 + 1.5] – 2
= 2[6.5] – 2
= 13 – 2
= 11 cm2
Question 5.
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth, and height of 15 m, 10 m, and 7 m respectively. From each can of paint 100 m2 of the area is painted. How many cans of paint will she need to paint the room?
Solution:
Length of hall = 15 m
The breadth of hall = 10 m
Height of hall = 7 m
Area of wall and ceiling of cuboidal hall = 2(l + b)h + lb
= 2[15 + 10] × 7 + 15 × 10
= 2[25] × 7 + 150
= 350 + 150
= 500 m2
1 m2 is painted by = 1 can
1 m2 is painted by = 1/100
500 m2 is painted by = 1/100 × 500 = 5 cans.
Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How many sheets of metal are required?
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