GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions 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?
GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3 img 1
Solution:
Total surface area of the cuboid (a)
= 2[lb + bh + hl]
= 2[(60 × 40) + (40 × 50) + (50 × 60)] cm2
= 2[2400 + 2000 + 3000] cm2
= 2[7400 cm2] = 14800 cm2

Total surface area of the cuboid (b)
= 2[lb + bh + hl]
= 2[(50 × 50) + (50 × 50) + (50 × 50)]
= 2[2500 + 2500 + 2500] cm2
= 2[7500 cm2] = 15000 cm2

Since the total surface area of the second (b) is more.
∴ Cuboid (a) will require lesser material.

GSEB Solutions 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 metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Total surface area of a suitcase
= 2 (lb + bh + hl)
= 2(80 × 48 + 48 × 24 + 24 × 80)
= 2(3840 + 1152 + 1920) cm2
= 2(6912) cm2
= 13824 cm2

∴ Total S.A. of 100 suitcases
= 100 × 13824 cm2
= 1382400 cm2
Surface area of 1 m of tarpaulin
= 100 cm × 96 cm = 9600 cm2
i.e., 1 m of tarpaulin can cover 9600 cm2 surface

∴ \(\frac{1382400}{9600}\) metres of tarpaulin will cover the total surfaces of 100 suitcases
i.e., \(\frac{1382400}{9600}\) or 144 metres of tarpaulin will be required to cover 100 suitcases.

Question 3.
Find the side of a cube whose surface area is 600 cm2.
Solution:
Let the side be x cm.
∴ Total S.A. of the cube = 6 × x2 cm2
But the S.A. of the cube = 600 cm2
∴ 6x2 = 600
or x2 = \(\frac{600}{6}\) = 100
or x2 = 102 ⇒ x = 10
∴ The required side of the cube = 10 cm

GSEB Solutions 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 bottom of the cabinet?
Solution:
Total surface area = 2[lb × bh + hl]
But bottom is not painted,
i.e., Area not painted = lb
∴ Total S.A. to be painted = 2(bh + hl) + lb
= 2[(2 × 1.5) + (1.5 × 1)] + (2 × 1) m2
= 2[3 + 1.5] + 2 m2
= 2[4.5] + 2 m2
= 9 + 2 m2
= 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 area is painted. How many cans of paint will she need to paint the room?
Solution:
Area to be painted
= [Area of 4-walls] + Area of ceiling
= 2[Perimeter × Height] + l × b
= 2[(15 + 10) × 7] + 15 × 10 m2
= 2[25 × 7] + 150 m2
= 2 × 175 + 150 m2
= 350 + 150 m2
= 500 m2
Since 1 can of paint covers 100 m2 area.
∴ Number of cans required = \(\frac{500}{100}\) = 5

GSEB Solutions 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 larger surface area?
GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3 img 2
Solution:
Similarity: Both have the same height.
Difference: Cylinder has curved and circular surfaces.
Cube: All faces are identical squares. Lateral surface area of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 7 cm2 = 7 × 22 cm2 = 154 cm2
Lateral surface area of the cube = 4l2
= 4 × (7 × 7) cm2
= 4 × 49 cm2
= 196 cm2
Obviously the cubical box has larger surface area.

Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution:
Here, r = 7 m and h = 3 m
∴ Total surface area = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + 3) m2
= 2 × \(\frac{22}{7}\) × 7 × 10 m2
= 2 × 22 × 10 m2 = 440 m2
Thus, 440 m2 metal is required.

GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3

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 rectangular sheet?
Solution:
∵ Lateral surface area of a cylinder
= 4224 cm2 and let length of the rectangular sheet = l cm
∴ 33 × l = 4224
∴ l = \(\frac{4224}{33}\) = 128 cm
Now perimeter of the rectangular sheet
= 2 [Length + Breadth]
= 2 [128 cm + 33 cm]
= 2 × 161 cm
= 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?
Solution:
The road roller is a cylinder.
Radius = \(\frac{84}{2}\) cm = 42 cm
GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3 img 3
Length (h) = 1 m = 100 cm
∴ Lateral surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 42 × 100 cm2
= 2 × 22 × 6 × 100 cm2 = 26400 cm2
∴ Area levelled by the roller in 1 revolution = 26400 cm2
or Area levelled in 750 revolutions
= 750 × 26400 cm2 = \(\frac{750×26400}{100×100}\) m2
= 15 × 132 m2 = 1980 m2
Thus, the required area of the road = 1980 m2

GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3

Question 10.
A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. 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?
GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3 img 4
Solution:
The label is in the shape of a cylinder having 14
Radius = \(\frac{14}{2}\) cm = 7 cm and Height = (2 + 2) cm less than the height of the container
or Height of the label = 20 cm – [2 + 2] cm = 16 cm
[∵ The label is placed 2 cm from top and bottom.]
∴ Area of the label = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 16 cm2
= 2 × 22 × 16 cm2 = 704 cm2

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