# Class 8 Maths NCERT Solutions for Chapter – 11 Mensuration Ex – 11.3

## Mensuration

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 first box
= 2(lb + bh + lh)
= 2 (60 x 40 + 40 x 50 + 60 x 50) cm2
= 200 (24+20+30) cm2
= 200 x 74 cm2
=14800 cm2
Total surface area of second box
= 6 (Edge)2 = 6 x 50 x 50 cm2
= 15000 cm2
Since the total surface area of first box is less than that of the second, therefore the first box i.e., (a) requires the least amount of material to make.

Question 2.
A suitcase of measures 80 cm x 48 cm x 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 the suitcase
= 2 (80 x 48 + 48 x 24 + 80 x 24) cm2
= 2(3840 + 1152 + 1920) cm2
= 2 x 6912 cm2 =13824 cm2
Area of 1 metre of trapaulin = (100 x 96) cm2
= 9600 cm2
Surface area of 100 suitcases =100 x 13824 cm2
Trapaulin required to cover 100 suitcases $\frac { 100\times 13824 }{ 9600 } meters$
= 144 metres

Question 3.
Find the side of a cube whose surface area is 600 cm2.

Solution:
Let a be the side of the cube having surface area 600 cm2.
∴ 6a2 =600 ⇒ a2 =100 ⇒ a =10
Hence, the side of the cube = 10 cm.

Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Solution:
Here l = 2 m, b = 1 m and h = 1.5 m
Area to be painted = 2bh + 2lh + lb
= (2 x 1 x 1.5 + 2 x 2 x 1.5 + 2 x 1) m2
= (3 + 6+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:
Here l = 15 m, b = 10 m and h = 7 m
Area to be painted = 2bh + 2lh + lb
= (2 x 10 x 7 + 2 x 15 x 7 + 15 x 10) m2
= (140 + 210 + 150) m2 =500 m2
Since each can of paint covers 100 m2, therefore 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 lateral surface area?

Solution:
The given figure are alike in their heights. They differ as under :
(i) One is named cylinder and the other is a cube.
(ii) Cylinder is a solid obtained by revolving a rectangular lamina about one of its sides whereas cube is solid bounded by six squares.
(iii) Cylinder has two circular faces, whereas cube has six faces.
Given,
r = $\frac { 7 }{ 2 }$ cm, h = 7 cm = $\left( 2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 22 } \times 7 \right)$ cm2 = 154 cm2
Lateral surface area of cylinder = 2πrh and, lateral surface area of cube
= Perimeter of base x height = (4 x 7 x 7) cm2 = 196 cm2
Clearly, cube has larger lateral 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.
Sheet of metal required to make a closed cylinder
= Total surface area of the cylinder
= (2πrh + 2πr2) sq. units $=\left( 2\times \frac { 22 }{ 7 } \times 7\times 3+2\times \frac { 22 }{ 7 } \times 7\times 7 \right)$ 2
=(138 + 308)m2 = 440 m2

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:
When a hollow cylinder is cut along its height and formed into a rectangular sheet. Then, the circumference of the base of cylinder is equal to the length of the sheet and the height of the cylinder is equal to the breadth of the sheet. Therefore, lateral surface area of the cylinder is equal to the area of the rectangular sheet. Let? be the length of the sheet.
∴ Area of the sheet = Lateral surface area of cylinder
⇒ l x 33 = 4224
⇒ $l=\left( \frac { 4224 }{ 33 } \right) cm=128cm$
Now, perimeter of rectangular sheet
= 2 (l + b) = 2 (128 + 33) cm
= 2 x 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: 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 surface area of the label.

Solution:

Since the company places a label around the surface of the cylindrical container of radius 7 cm and height 20 cm such that it is placed 2 cm from top and bottom.
∴ We have to find the curved surface of a cylinder of radius 7 cm and height (20 – 4) cm i.e., 16 cm. ,
This curved surface area = $\left( 2\times \frac { 22 }{ 7 } \times 7\times 16 \right) { cm }^{ 2 }$
= 704 cm2