Exercise 11.3
Q 1 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
Question 1
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Sol :
We know that,
Total surface area of the cuboid = 2 (lh + bh + lb)
Total surface area of the cube = 6 (l)2
Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2
= [2(2400 + 2000 + 3000)] cm2
= (2 × 7400) cm2
= 14800 cm2
Total surface area of cube (b) = 6 (50 cm)2 = 15000 cm2
Thus, the cuboidal box (a) will require lesser amount of material.
Q 2 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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?
Sol :
Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80)]
= 2[3840 + 1152 + 1920]
= 13824 cm2
Total surface area of 100 suitcases = (13824 × 100) cm2 = 1382400 cm2
Required tarpaulin = Length × Breadth
1382400 cm2 = Length × 96 cm
Length =
= 14400 cm = 144 mThus, 144 m of tarpaulin is required to cover 100 suitcases.
Q 3 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
Find the side of a cube whose surface area is 600 cm2.
Sol :
Given that, surface area of cube = 600 cm2
Let the length of each side of cube be l.
Surface area of cube = 6 (Side)2
600 cm2 = 6l2
l2= 100 cm2
l = 10 cm
Thus, the side of the cube is 10 cm.
Q 4 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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 the bottom of the cabinet?
Sol :
Length (l) of the cabinet = 2 m
Breadth (b) of the cabinet = 1 m
Height (h) of the cabinet = 1.5 m
Area of the cabinet that was painted = 2h (l + b) + lb
= [2 × 1.5 × (2 + 1) + (2) (1)] m2
= [3(3) + 2] m2
= (9 + 2) m2
= 11 m2
Q 5 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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?
Sol :
Given that,
Length (l) = 15 m, breadth (b) = 10 m, height (h) = 7 m
Area of the hall to be painted = Area of the wall + Area of the ceiling
= 2h (l + b) + lb
= [2(7) (15 + 10) + 15 ×10] m2
= [14(25) + 150] m2
= 500 m2
It is given that 100 m2 area can be painted from each can.
Number of cans required to paint an area of 500 m2
=
Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.
Q 6 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Sol :
Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Q 7 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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?
Sol :
Total surface area of cylinder = 2πr (r + h)
m2= 440 m2
Thus, 440 m2 sheet of metal is required.
Q 8 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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?
Sol :
A hollow cylinder is cut along its height to form a rectangular sheet.
Area of cylinder = Area of rectangular sheet
4224 cm2 = 33 cm × Length
Thus, the length of the rectangular sheet is 128 cm.
Perimeter of the rectangular sheet = 2 (Length + Width)
= [2 (128 + 33)] cm
= (2 × 161) cm
= 322 cm
Q 9 - Ex 11.3 - Mensuration - NCERT Maths Class 8th - Chapter 11
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.
Sol :
In one revolution, the roller will cover an area equal to its lateral surface area.
Thus, in 1 revolution, area of the road covered = 2πrh
In 750 revolutions, area of the road covered
=
= 1980 m2
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.
Sol :
Height of the label = 20 cm − 2 cm − 2 cm = 16 cm
Radius of the label
Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm.
Area of the label = 2π (Radius) (Height)
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