Class 8 NCERT Solutions – 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? 

Solution:

(a) For finding which box requires the lesser amount of material to make we have to calculate total surface area of both these boxes. 

So, Length of first cuboidal box (l) = 60 cm

Breadth of first cuboidal box (b) = 40 cm

Height of first cuboidal box (h) = 50 cm

Total surface area of first cuboidal box = 2 × (lb + bh + hl)

= 2 × (60 × 40 + 40 × 50 + 50 × 60)

= 2 × (2400 + 2000 + 3000)

= 14800 cm²

Total surface area of first cuboidal box is 14800 cm².

(b) Also, 

Length of second cubical box (l) = 50 cm

Breadth of second cubicalbox (b) = 50 cm

Height of second cubicalbox (h) = 50 cm

Total surface area of second cubical box = 6 (side)²

= 6 (50 × 50)

= 6 × 2500

= 15000 cm²

Total surface area of the second cubical box is 15000 cm²

From both of these result we found that cuboidal box (observation “a”) requires the lesser amount of material to make.

Question 2. A suitcase measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases? 

Solution:

 Length of suitcase box (l)= 80 cm,  

Breadth of suitcase box (b) = 48 cm

Height of cuboidal box (h) = 24 cm

Total surface area of suitcase box = 2 × (lb + bh + hl)  

= 2 × (80 × 48 + 48 × 24 + 24 × 80)

= 2 × (3840 + 1152 + 1920)

= 2 × 6912

= 13824 cm²

Hence, Total surface area of suitcase box is 13824 cm²

As suitcase is fully covered by tarpaulin

Area of Tarpaulin cloth = Surface area of suitcase

(l × b) = 13824

(l × 96) = 13824

l = 144m

Required tarpaulin for 100 suitcases = 144 × 100 = 14400 cm = 144 m

Hence, tarpaulin cloth required to cover 100 suitcases is 144 m.

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

Solution:

Given that

Surface area of cube = 600 cm²

Formula for surface area of a cube = 6(side)²

Substituting the values, we get

6 × (side)² = 600

(side)² = 100

Or side = ±10

We can’t take side as negative

Hence, the measure of each side of a cube is 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 the bottom of the cabinet. 

Solution:

Length of cabinet (l) = 2 m, 

Breadth of cabinet (b) = 1 m, 

and Height of cabinet (h) = 1.5 m

Surface area of cabinet = 2 × (lb +  bh + hl) – lb

= 2 × (2 × 1 + 1 × 1.5 + 1.5 × 2)  – 2 × 1

= 2 × (2 +  1.5 + 3.0) – 2

= 2 × (6.5) – 2

= 13 – 2

= 11m²

Required surface area of cabinet is 11m²

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:

Length of wall (l) = 15 m, 

Breadth of wall (b) = 10 m 

Height of wall, (h) = 7 m

Total Surface area of classroom = 2 × (lb + bh + hl) – lb

= 2 × (15 × 10 +  10 × 7 + 7 × 15) – (15 × 10) 

= 2 × (150 + 70 + 105) – 150

= 650 – 150

= 500

Now, Required number of cans = Area of hall/Area of one can  

= 500/100 = 5

Therefore, 5 cans are required to paint the room.

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

Solution:

Given that,

Diameter of cylinder = 7 cm 

Radius of cylinder (r) = 7 / 2 cm

Height of cylinder (h) = 7 cm

Surface area of cylinder = 2πrh  

= 2 × (22 / 7) × (7 / 2) × 7 = 154

So, Lateral surface area of cylinder is 154 cm²

Now, lateral surface area of cube = 4 (side)² = 4 × 72 = 4 × 49 = 196

Lateral surface area of cube is 196 cm²

Hence, the 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 many sheets of metal is required? 

Radius of cylindrical tank (r) = 7 m

Height of cylindrical tank (h) = 3 m

Total surface area of cylindrical tank = 2πrh + 2πr² = 2πr × (h + r)

= 2 × (22 / 7) × 7 (3 + 7)  

= 44 × 10 = 440

Therefore, 440 m² metal sheet 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 the rectangular sheet?

Solution: 

Lateral surface area of hollow cylinder = 4224 cm²

Height of hollow cylinder, h = 33 cm

and say r be the radius of the hollow cylinder

Curved surface area of hollow cylinder = 2πrh  

4224 = 2 × π × r × 33

r = (4224) / (2π × 33)

r = 64 × 7/ 22

Now, Length of rectangular sheet, l = 2πr  

l = 2 × 22 / 7× (64 × 7/ 22) = 128 

So the length of the rectangular sheet is 128 cm.

Also, Perimeter of rectangular sheet = 2 × (l + b)  

= 2 × (128 + 33)  

= 322

The perimeter of rectangular sheet is 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:

Diameter of road roller (d) = 84 cm

Radius of road roller (r) = d / 2 = 84 / 2 = 42 cm

Length of road roller (h) = 1 m = 100 cm

Curved surface area of road roller = 2πrh  

= 2 × (22 / 7) × 42 × 100 = 26400

Curved surface area of road roller is 26400 cm²

Again, Area covered by road roller in 750 revolutions = 26400 × 750cm²

= 1,98,00,000 cm²

= 1,98,00,000 × (1 / 10,000) m² (1cm²= 1 / 10,000m²) 

Hence, the area of the road is 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:

Diameter of cylindrical container (d) = 14 cm

Radius of cylindrical container (r) = d / 2 = 14 / 2 = 7 cm

Height of cylindrical container (H) = 20 cm

Height of the label (h) = Height of container – free space

= 20 – 2 (2) 

= 16 cm

Area of label = 2πrh

= 2 × (22 / 7) × 7 × 16

= 704 cm²

Hence, the area of the label is 704 cm².