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)] cm^{2}

= 2[2400 + 2000 + 3000] cm^{2}

= 2[7400 cm^{2}] = 14800 cm^{2}

Total surface area of the cuboid (b)

= 2[lb + bh + hl]

= 2[(50 × 50) + (50 × 50) + (50 × 50)]

= 2[2500 + 2500 + 2500] cm^{2}

= 2[7500 cm^{2}] = 15000 cm^{2}

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) cm^{2}

= 2(6912) cm^{2}

= 13824 cm^{2}

∴ Total S.A. of 100 suitcases

= 100 × 13824 cm^{2}

= 1382400 cm^{2}

Surface area of 1 m of tarpaulin

= 100 cm × 96 cm = 9600 cm^{2}

i.e., 1 m of tarpaulin can cover 9600 cm^{2} 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 cm^{2}.

Solution:

Let the side be x cm.

∴ Total S.A. of the cube = 6 × x^{2} cm^{2}

But the S.A. of the cube = 600 cm^{2}

∴ 6x^{2} = 600

or x^{2} = \(\frac{600}{6}\) = 100

or x^{2} = 10^{2} ⇒ 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) m^{2}

= 2[3 + 1.5] + 2 m^{2}

= 2[4.5] + 2 m^{2}

= 9 + 2 m^{2}

= 11 m^{2}

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 m^{2} 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 m^{2}

= 2[25 × 7] + 150 m^{2}

= 2 × 175 + 150 m^{2}

= 350 + 150 m^{2}

= 500 m^{2}

Since 1 can of paint covers 100 m^{2} 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 cm^{2} = 7 × 22 cm^{2} = 154 cm^{2}

Lateral surface area of the cube = 4l^{2}

= 4 × (7 × 7) cm^{2}

= 4 × 49 cm^{2}

= 196 cm^{2}

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) m^{2}

= 2 × \(\frac{22}{7}\) × 7 × 10 m^{2}

= 2 × 22 × 10 m^{2} = 440 m^{2}

Thus, 440 m^{2} metal is required.

Question 8.

The lateral surface area of a hollow cylinder is 4224 cm^{2}. 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 cm^{2} 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 cm^{2}

= 2 × 22 × 6 × 100 cm^{2} = 26400 cm^{2}

∴ Area levelled by the roller in 1 revolution = 26400 cm^{2}

or Area levelled in 750 revolutions

= 750 × 26400 cm^{2} = \(\frac{750×26400}{100×100}\) m^{2}

= 15 × 132 m^{2} = 1980 m^{2}

Thus, the required area of the road = 1980 m^{2}

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 cm^{2}

= 2 × 22 × 16 cm^{2} = 704 cm^{2}