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

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 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?

(a) Here, l = 60 cm, b = 40 cm; h = 50 cm Total surface area of the cuboid
= 2 (lb + bh + hl)
= 2 [(60 × 40) + (40 × 50) + (50 × 60)] cm2
= 2 [2400 + 2000 + 3000] cm2
= 2 (7400) cm2 = 14800 cm2

(b) Here, l = 50 cm, b = 50 cm, h = 50 cm
Total surface area of the cuboid
= 2 (lb + bh + lh)
= 2 [(50 × 50)+ (50 × 50)+ (50 × 50)] cm2
= 2 [2500 + 2500 + 2500] cm2
= 2 × 7500 cm2 = 15000 cm2
Since, the total surface area of the second (b) is greater:
∴ Cuboid (a) will required 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?
Here, length = 80 cm, breadth = 48 cm, height = 24 cm
Total surface are a of asuitcase = 2(lb+bh+lh)
= 2[80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]cm2
= 2[6912] cm2 = 13824 cm2
Required tarpaulin for 100 suitcases = 13824 x 100 cm2
Width of the tarpaulin = 96 cm
Length, tarpaulin required = $$\frac{13824 \times 100}{96}$$cm
[1 m = 100 cm]
= $$\frac{13824 \times 100}{96 \times 100}$$m = $$\frac{13824}{96}$$m = 144 m
The required length of tarpaulin for 100 suitcases = 144 m

Question 3.
Find the side of a cube whose surface area is 600 cm2
Let the side of a cube be ‘x’ cm
Total surface area of the cube = 600 cm2
6 × x2 = 600
x2 = $$\frac{600}{6}$$
x2 = 100
x2 = 102
x = 10
The required side of the cube =10 cm.

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

Total surface area of the cabinet = 2 × [lb + bh + lh] sq m
Area not to be painted = Bottom of the cabinet = lb
Area to be painted = T.S.A – lb
= 2 (lb + bh + lh) – lb
= 2[1 × 2 + 2 × 1.5 + 1.5 × 1]-(1 × 2)m2
= 2(2 + 3 + 1.5) -2 m2
= 2 × 6.5 – 2 m2
= 13 – 2 m2
= 11 m2
Area to be painted = 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?
Here l = 15 m, b = 10 m and h = 7 m.
Area to the painted = Area of four walls + Area of ceiling.
= 2 (bh + hl) + lb
2 [10 × 7 +7 × 15] + 15 × 10 m2
= 2 [70 + 105] + 150 m2
= 2(175) + 150 m2
= 350 + 150 m2 = 500 m2
= Number of cans needed = $$\frac{\text { Area to be painted }}{\text { Area painted by } 1 \text { can }}$$ = $$\frac{500}{100}$$ = 5
Number of cans needed = 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?

Similarity: Both have the same height
Difference: Cylinder has curved and circular surfaces.
but in 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 sq.m
= 4 × 7 × 7 cm2 = 196 cm2

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?
Here r = 7m, h = 3m
Total surface area of the cylinder = 2 π r (r + h)sq. units
= 2 × $$\frac{22}{7}$$ × 7(7 + 3) = 44 × 10 m2 = 440 m2
Metal required for the tank = 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?
Lateral surface area of the cylinder = 4224 cm2
Let the length of the rectangular sheet be l cm
Width of the sheet = 33 cm.
Area of the rectangular sheet = L.S.A of the cylinder
l × 33 = 4224
l = $$\frac{4224}{33}$$ cm = 128 cm
Perimeter of the rectangular sheet = 2 (l + b)
= 2 (128 + 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 1m.

Radius of the roller = $$\frac{84}{2}$$ = 42cm
Length of the roller = 1 m = 100 cm
Lateral surface area = 2 π rh sq unit
= 2 × $$\frac{22}{7}$$ × 42 × 100 cm2
= 44 × 6 × 100 cm2 = 26400 cm2
Area levelled by 750 revolutions
= 750 × 26400 cm2
=$$\frac{750 \times 26400}{100 \times 100}$$m2 = $$\frac{75 \times 264}{10}$$ = 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.

Radius of the label = $$\frac{14}{2}$$ = 7 cm
= 2πrh = 2 × $$\frac{22}{7}$$ × 7 × 16