# 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 Ã— (lb + bh + hl)

= 2 Ã— (60 Ã— 40 + 40 Ã— 50 + 50 Ã— 60)

= 2 Ã— (2400 + 2000 + 3000)

= 14800 cm

^{2}Total surface area of first cuboidal box is 14800 cmÂ².

(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 Ã— 50)

= 6 Ã— 2500

= 15000 cmÂ²

Total surface area of the second cubical box is 15000 cm

^{2}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 Ã— 48 cm Ã— 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 Ã— (lb + bh + hl)

= 2 Ã— (80 Ã— 48 + 48 Ã— 24 + 24 Ã— 80)

= 2 Ã— (3840 + 1152 + 1920)

= 2 Ã— 6912

= 13824 cmÂ²

Hence, Total surface area of suitcase box is 13824 cmÂ²

As suitcase is fully covered by tarpaulin

Area of Tarpaulin cloth = Surface area of suitcase

(l Ã— b) = 13824

(l Ã— 96) = 13824

l = 144m

Required tarpaulin for 100 suitcases = 144 Ã— 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 cmÂ².**

**Solution:**

Given that

Surface area of cube = 600 cmÂ²

Formula for surface area of a cube = 6(side)Â²

Substituting the values, we get

6 Ã— (side)Â² = 600

(side)Â² = 100

Or side = Â±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 Ã— 2 m Ã— 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 Ã— (lb + bh + hl) – lb

= 2 Ã— (2 Ã— 1 + 1 Ã— 1.5 + 1.5 Ã— 2) – 2 Ã— 1

= 2 Ã— (2 + 1.5 + 3.0) – 2

= 2 Ã— (6.5) – 2

= 13 – 2

= 11mÂ²

Required surface area of cabinet is 11mÂ²

**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 mÂ² 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 Ã— (lb + bh + hl) – lb

= 2 Ã— (15 Ã— 10 + 10 Ã— 7 + 7 Ã— 15) – (15 Ã— 10)

= 2 Ã— (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 = 2Ï€rh

= 2 Ã— (22 / 7) Ã— (7 / 2) Ã— 7 = 154

So, Lateral surface area of cylinder is 154 cmÂ²

Now, lateral surface area of cube = 4 (side)Â² = 4 Ã— 72 = 4 Ã— 49 = 196

Lateral surface area of cube is 196 cmÂ²

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 = 2Ï€rh + 2Ï€rÂ² = 2Ï€r Ã— (h + r)

= 2 Ã— (22 / 7) Ã— 7 (3 + 7)

= 44 Ã— 10 = 440

Therefore, 440 mÂ² metal sheet is required.

**Question 8. The lateral surface area of a hollow cylinder is 4224 cmÂ² 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 cmÂ²

Height of hollow cylinder, h = 33 cm

and say r be the radius of the hollow cylinder

Curved surface area of hollow cylinder = 2Ï€rh

4224 = 2 Ã— Ï€ Ã— r Ã— 33

r = (4224) / (2Ï€ Ã— 33)

r = 64 Ã— 7/ 22

Now, Length of rectangular sheet, l = 2Ï€r

l = 2 Ã— 22 / 7Ã— (64 Ã— 7/ 22) = 128

So the length of the rectangular sheet is 128 cm.

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

= 2 Ã— (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 = 2Ï€rh

= 2 Ã— (22 / 7) Ã— 42 Ã— 100 = 26400

Curved surface area of road roller is 26400 cmÂ²

Again, Area covered by road roller in 750 revolutions = 26400 Ã— 750cmÂ²

= 1,98,00,000 cmÂ²

= 1,98,00,000 Ã— (1 / 10,000) mÂ² (1cmÂ²= 1 / 10,000mÂ²)

Hence, the area of the road is 1980 mÂ².

**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 â€“ 2 (2)

= 16 cm

Area of label = 2Ï€rh

= 2 Ã— (22 / 7) Ã— 7 Ã— 16

= 704 cmÂ²

Hence, the area of the label is 704 cmÂ².

## Please

Loginto comment...