# NCERT Solutions for Class 8 Maths Exercise 11.3

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NCERT solutions for Class 8 Maths Mensuration

## NCERT Solutions for Class 8 Maths Mensuration

###### 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Ans. (a) Given: Length of cuboidal box  = 60 cm

Breadth of cuboidal box  = 40 cm

Height of cuboidal box  = 50 cm

Total surface area of cuboidal box

= ]

= 2 (60  40 + 40  50 + 50 60)

= 2 (2400 + 2000 + 3000

= 2  7400 = 14800

(b) Given: Length of cuboidal box

= 50 cm

Breadth of cuboidal box  = 50 cm

Height of cuboidal box  = 50 cm

Total surface area of cuboidal box

=

= 2 (50  50 + 50  50 + 50  50)

= 2 (2500 + 2500 + 2500)

= 2  7500 = 15000

Hence cuboidal box  requires the lesser amount of materal to make, since surface area of box  is less than that of box

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 2. A suitcase with 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?

Ans. Given: Length of suitcase box = 80 cm, Breadth of suitcase box  = 48 cm

And Height of cuboidal box  = 24 cm

Total surface area of suitcase box

=

= 2 (80  48 + 48  24 + 24  80)

= 2 (3840 + 1152 + 1920)

= 2  6912 = 13824

Area of Tarpaulin cloth = Surface area of suitcase

= 13824

= 144 cm

Required tarpaulin for 100 suitcases

= 144  100 = 14400 cm = 144 m

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

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 3. Find the side of a cube whose surface area id 600.

Ans. Here Surface area of cube = 600 cm2

= 600

= 100

= 10 cm

Hence the side of cube is 10 cm

###### 4. Rukshar 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?

Ans. Here, Length of cabinet  = 2 m, Breadth of cabinet  = 1 m

And Height of cabinet  = 1.5 m

Surface area of cabinet =

= 2  1 + 2 (1  1.5 + 1.5  2)

= 2 + 2 (1.5 + 3.0)

= 2 + 9.0

= 11

Hence required surface area of cabinet is 11.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 5. Daniel is paining 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 of area is painted. How many cans of paint will she need to paint the room?

Ans. Here, Length of wall  = 15 m, Breadth of wall  = 10 m

And Height of wall  = 7 m

Total Surface area of classroom

=

= 15  10 + 2 (10  7 + 7 15)

= 150 + 2 (70 + 105)

= 150 + 350

= 500

Now Required number of cans

=  = 5 cans

Hence 5 cans are required to paint the room.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?

Ans. Given: Diameter of cylinder = 7 cm

And Height of cylinder  = 7 cm

Lateral surface area of cylinder =

=

= 154 cm2

Now lateral surface area of cube

=  = 4  49 = 196

Hence the cube has larger lateral surface area.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 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?

Ans. Given: Radius of cylindrical tank  = 7 m

Height of cylindrical tank  = 3 m

Total surface area of cylindrical tank

=

=

= 44  10 = 440

Hence 440  metal sheet is required.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 8. The lateral surface area of a hollow cylinder is 4224. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Ans. Given: Lateral surface area of hollow cylinder = 4224

And Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder =

4224 =

=  cm

Now Length of rectangular sheet =

= 128 cm

Perimeter of rectangular sheet =

= 2 (128 + 33) = 2 x 161 = 322 cm

Hence perimeter of rectangular sheet is 322 cm.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 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 1 m.

Ans. Given: Diameter of road roller = 84 cm

= 42 cm

Length of road roller  = 1 m = 100 cm

Curved surface area of road roller =  =  = 26400 cm2

Area covered by road roller in 750 revolutions = 26400  750

= 1,98,00,000

= 1980 m2 [ 1 = 10,000]

Hence the area of the road is 1980.

NCERT Solutions for Class 8 Maths Exercise 11.3

###### 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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Ans. Given: Diameter of cylindrical container = 14 cm

Radius of cylindrical container  = 7 cm

Height of cylindrical container = 20 cm

Height of the label = 20 – 2 – 2

= 16 cm

Curved surface area of label =

=  = 704 cm2

Hence the area of the label of 704 cm2.

## NCERT Solutions for Class 8 Maths Exercise 11.3

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### 23 thoughts on “NCERT Solutions for Class 8 Maths Exercise 11.3”

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