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?

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.

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

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

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has larger surface area?

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.

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

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

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?

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