You can Download KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers to help you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 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) Length of cuboid (l) = 60 cm

Breadth of cuboid (b) = 40 cm

Height of cuboil (h) = 50 cm

Anount of material required = Total surface area of cuboid

= 2(lb + bh + lh)

= 2(60 × 40 + 40 × 50 + 60 × 50)

= 2[2400 + 2000 + 3000]

= 2 × 7400

= 14800 cm^{2}

Side of cube = 50 cm

(b) Total material required = Total surface area of cube

= 6(side)^{2}

= 6(50)^{2}

= 6 × 2500

= 15000 cm^{2}

Cuboidal boxes required a lesser amount of material to make.

Question 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?

Solution:

Measurement of suitcase = 80 cm × 48 cm × 24 cm

Surface area of suitcase = 2(lb + bh + lh)

= 2[80 × 48 + 48 × 24 + 80 × 24]

= 2[3840 + 1152 + 1920]

= 2[6912]

= 13824 cm^{2}

Area of cloth required for 100 suitcases = 100 × 13824 = 1382400 cm^{2}

Area of cloth = l × b = l × 96

l × 96 = 1382400

l = \(\frac{1382400}{96}\) = 14400 cm (approx.)

or length of cloth = \(\frac{14400}{100}\) m = 144 mts.

Question 3.

Find the side of a cube whose surface area is 600 cm^{2}.

Solution:

Surface area of cube = 600 cm^{2}

⇒ 6(Side)^{2} = 600

⇒ (Side)^{2} = \(\frac{600}{6}\) = 100

⇒ Side = √100 = 10 mts

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:

Measurement of cabinet = 1 m × 2 m × 1.5 m

Surface area of cabinet except base = 2 (lb + bh + lh) – lb

= 2[1 × 2 + 2 × 1.5 + 1.5 × 1] – 1 × 2

= 2[2 + 3 + 1.5] – 2

= 2[6.5] – 2

= 13 – 2

= 11 cm^{2}

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 m^{2} of the area is painted. How many cans of paint will she need to paint the room?

Solution:

Length of hall = 15 m

The breadth of hall = 10 m

Height of hall = 7 m

Area of wall and ceiling of cuboidal hall = 2(l + b)h + lb

= 2[15 + 10] × 7 + 15 × 10

= 2[25] × 7 + 150

= 350 + 150

= 500 m^{2}

1 m^{2} is painted by = 1 can

1 m^{2} is painted by = \(\frac{1}{100}\)

500 m^{2} is painted by = \(\frac{1}{100}\) × 500 = 5 cans.

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:

They are alike, both have the same height. They are different as one is cylindrical in shape and other is cubical.

Lateral surface area of cylinder = 2πrh

= \(2 \times \frac{22}{7} \times \frac{7}{2} \times 7\)

= 154 cm^{2}

Lateral surface area of cube = 4(side)^{2}

= 4(7)^{2}

= 4 × 49

= 196 cm^{2}

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 are required?

Solution:

Radius of cylinder = 7 cm

Height of cylinder = 3 m

Total surface area of cylinder = 2πr(r + h)

= 2 × \(\frac{22}{7}\) × 7[7 + 3]

= 44 × 10

= 440 m^{2}

Hence, 440 m^{2} of metal sheet is required to make the closed cylindrical tank.

Question 8.

The lateral surface area of a hollow cylinder 4224 cm^{2}. 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 cm^{2}

Width of rectangular sheet = 33 cm

Area of sheet = LSA of cylinder

⇒ l × b = 4224

⇒ l × 33 = 4224

⇒ l = \(\frac{4224}{33}\) = 128 cm

Perimeter of rectangular sheet = 2(l + b)

= 2(128 + 33)

= 2 × 161

= 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 the length is 1 m.

Solution:

Length of roller = 1 m

Radius of roller = \(\frac{84}{2}\) = 42 cm or \(\frac{42}{100}\) = 0.42 m.

Area covered in 1 revolution = 2πrh

= 2 × \(\frac{22}{7}\) × 0.42 × 1

= 44 × 0.06

= 2.64 m^{2}

Area covered in 750 revolution = 750 × 2.64 = 1980 m^{2}

∴ Area of road = 1980 m^{2}

Question 10.

A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height of 20 cm. The 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:

Height of cylindrical container = 20 cm

Label is placed 2 cm from top and bottom.

Height of label = 20 – (2 + 2)

= 20 – 4

= 16 cm

Radius of cylinder = 7 cm

Area of label = 2πrh

= 2 × \(\frac{22}{7}\) × 7 × 16

= 44 × 16

= 704 cm^{2}