Chapter 11 Mensuration Exercise 11.3

Question 1.
There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q1
Solution:
(a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 120000 cm3
(b) Volume of cube = (Side)3 = (50)3 = 50 × 50 × 50 = 125000 cm3
Cuboidal box (a) requires lesser amount of material.

Question 2.
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of the suitcase = 80 cm × 48 cm × 24 cm
l = 80 cm, b = 48 cm and h = 24 cm
Total surface area of the suitcase = 2[lb + bh + hl]
= 2 [80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]
= 2 × 6912
= 13824 cm2
Area of tarpaulin = length × breadth = l × 96 = 96l cm2
Area of tarpaulin = Area of 100 suitcase
96l = 100 × 13824
l = 100 × 144 = 14400 cm = 144 m
Hence, the required length of the cloth = 144 m.

Question 3.
Find the side of a cube whose surface area is 600 cm2?
Solution:
Total surface area of a cube = 6l2
6l2 = 600
l2 = 100
l = √100 = 10 cm
Hence, the required length of side = 10 cm.

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?
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q4
Solution:
l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box
= 2 [lb + bh + hl] – lb
= 2[2 × 1.5 + 1.5 × 1 + 1 × 2] – 2 × 1
= 2[3 + 1.5 + 2] – 2
= 2[6.5] – 2
= 13 – 2
= 11 m2
Hence, the required area = 11 m2.

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:
Surface area of a cuboidal hall without bottom = Total surface area – Area of base
= 2 [lb + bh + hl] – lb
= 2 [15 × 10 + 10 × 7 + 7 × 15] – 15 × 10
= 2[150 + 70 + 105] – 150
= 2 [325] – 150
= 650 – 150
= 500 m2
Area of the paint in one can = 100 m2

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q6

Question 8.
The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.
Solution:
Width of the rectangular sheet = Circumference of the cylinder
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q8
h = 128 cm
l = 128 cm, b = 33 cm
Perimeter of the sheet = 2(l + b) = 2(128 + 33) = 2 × 161 = 322 cm
Hence, the required perimeter = 322 cm.

Question 9.
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q9
Solution:
The lateral surface area of the road roller = 2πrh
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q9.1
Hence, the area of road = 1980 m2

Question 10.
A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q10

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-1
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-2
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-3
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-4
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-5
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-6
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 q-7

Chapter 11 Mensuration Exercise 11.3