KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

You can Download KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers to help you to revise the complete syllabus.

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?
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q1
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.

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

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 = \(\frac{1382400}{96}\) = 14400 cm (approx.)
or length of cloth = \(\frac{14400}{100}\) 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 = \(\frac{600}{6}\) = 100
⇒ Side = √100 = 10 mts

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

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.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q4
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 = \(\frac{1}{100}\)
500 m2 is painted by = \(\frac{1}{100}\) × 500 = 5 cans.

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

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?
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q6
Solution:
They are alike, both have the same height. They are different as one is cylindrical in shape and other is cubical.
Lateral surface area of cylinder = 2πrh
= \(2 \times \frac{22}{7} \times \frac{7}{2} \times 7\)
= 154 cm2
Lateral surface area of cube = 4(side)2
= 4(7)2
= 4 × 49
= 196 cm2

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?
Solution:
Radius of cylinder = 7 cm
Height of cylinder = 3 m
Total surface area of cylinder = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 7[7 + 3]
= 44 × 10
= 440 m2
Hence, 440 m2 of metal sheet is required to make the closed cylindrical tank.

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

Question 8.
The lateral surface area of a hollow cylinder 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:
Lateral surface area of hollow cylinder = 4224 cm2
Width of rectangular sheet = 33 cm
Area of sheet = LSA of cylinder
⇒ l × b = 4224
⇒ l × 33 = 4224
⇒ l = \(\frac{4224}{33}\) = 128 cm
Perimeter of rectangular sheet = 2(l + b)
= 2(128 + 33)
= 2 × 161
= 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 the length is 1 m.
Solution:
Length of roller = 1 m
Radius of roller = \(\frac{84}{2}\) = 42 cm or \(\frac{42}{100}\) = 0.42 m.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q9
Area covered in 1 revolution = 2πrh
= 2 × \(\frac{22}{7}\) × 0.42 × 1
= 44 × 0.06
= 2.64 m2
Area covered in 750 revolution = 750 × 2.64 = 1980 m2
∴ Area of road = 1980 m2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

Question 10.
A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height of 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?
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Q10
Solution:
Height of cylindrical container = 20 cm
Label is placed 2 cm from top and bottom.
Height of label = 20 – (2 + 2)
= 20 – 4
= 16 cm
Radius of cylinder = 7 cm
Area of label = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 16
= 44 × 16
= 704 cm2