## 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?

Answer:

Given

2 cuboidal boxes

Box A

l = 60 cm

b = 40 cm

h = 50 cm

Box B

l = 50 cm

b = 50 cm

h = 50 cm

To find

Which box requires lesser amount of material to make?

Formula

TSA of cuboid (Box A)= 2(lh + bh + lb)

TSA of cube (Box B) = 6s²

Working

Box A

= 2[(60)(50) + (40)(50) + (60)(40)]

= 2[3000 + 2000 + 2400]

= 2[7400]

= 14800 cm²

Box B

= 6 × (50)²

= 6 × 2500

= 15000 cm²

Comparing

15000 cm² > 14800 cm²

Therefore, Box A requires 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 metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Answer:

Given

Number of suitcases = 100

Length of suitcase = 80 cm

Breadth of suitcase = 48 cm

Height of suitcase = 24 cm

Width of tarpaulin cloth = 96 cm

To find

Length of cloth required to cover 100 suitcases

Formula

TSA = 2(lh + bh + lb)

Area of cloth = l × b

Working

= 2[(80)(24) + (48)(24) + (80)(48)]

= 2[1920 + 1152 + 3840]

= 2[6912]

= 13824 cm²

Length of cloth required for 1 suitcase = ?

= 13824 = l × 96

= l = 13824/96

= l = 144 cm

Length of cloth required for 100 suitcase = ?

= 144 × 100

= 14400 cm

= 144 m

Therefore, 144 metres of tarpaulin of width 96 cm is required to cover 100 such suitcases

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

Answer:

Given

TSA of cube = 600 cm²

To find

Side of the cube

Formula

TSA of cube = 6s²

Working

= 600 = 6s²

= s² = 600/6

= s² = 100

= √100

= s = 10 cm

Therefore, the side of cube whose surface area is 600 cm² 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?

Answer:

Given

Cuboidal cabinet

length = 1 m

breadth = 2 m

height = 1.5 m

To find

Surface area to be painted except the bottom

Formula

LSA + area of top = 2h(l + b) + lb

or

TSA - area of bottom

2(lh + bh + lb) - lb

Working

= 2 × 1.5(1+ 2) + 1 × 2

= 3 × 3 + 2

= 11 m²

Therefore, 11 m² of cabinet she has painted all except the bottom of the cabinet.

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?

Answer:

Given

Dimensions of room

length = 15 m

breadth = 10 m

height = 7 m

Area painted by each can = 100 m²

To find

Number of paint cans used

Formula

LSA + area of ceiling = 2h(l + b) + lb

or

TSA - area of floor = 2(lh + bh + lb) - lb

Working

= 2 × 7(15 + 10) + 15 × 10

= 14 × 25 + 150

= 350 + 150

= 500 m²

Number of cans required = ?

= 500/100

= 5 cans

Therefore, she would need 5 paint cans to paint the room.

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

Answer:

Similarities

→ Both have the same height.

Differences

→ The first one is a cylinder while the second one is cube.

→ Both have a different LSA. The cube has a larger one.

Cylinder LSA = 2πrh

= 2 × 22/7 × 3.5 × 7

= 2 × 22 × 3.5

= 154 cm²

Cube LSA = 4s²

= 4 × 7²

= 4 × 49

= 196 cm²

Question 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?

Answer:

Given

Closed cylindrical tank

radius = 7 m

height = 3 m

To find

Sheet of metal required to make the tank

Formula

TSA of cylinder = 2πr(h + r)

Working

= 2 × 22/7 × 7 × (3 + 7)

= 2 × 22 × 10

= 440 m²

Therefore, the sheet of metal required to make the closed cylindrical tank is 440 m².

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?

Answer:

Given

Hollow cylinder LSA = 4224 cm²

Width of rectangular sheet = 33 cm

To find

Perimeter of the rectangular sheet

Formula

LSA of cylinder = 2πrh

Area of rectangle = l × b

Perimeter of rectangle = 2(l + b)

Working

LSA of cylinder = area of rectangle

4224 = l × 33

l = 4224/33

l = 128 cm

Perimeter of rectangle = ?

= 2(l + b)

= 2(128 + 33)

= 2 × 161

= 322 cm

Therefore, 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.

Answer:

Given

Road roller

diameter = 84 cm

height = 1 m (100 cm)

Number of revolutions = 750

To find

The area of road

Formula

LSA of cylinder = 2πrh

Working

Area of road = ?

= 750 × 2πrh

= 750 × 2 × 22/7 × 42 × 100

= 750 × 2 × 22 × 6 × 100

= 19800000 cm²

(1 m² = 10000 cm²)

= 19800000/10000

= 1980 m²

Therefore, 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.

Answer:

Given

Cylindrical container

diameter = 14 cm

height = 20 cm

Label is placed 2 cm from top and bottom

To find

Area of label

Formula

LSA of cylinder = 2πrh

Working

(height of label is 16 cm because 2 cm from top and bottom are excluded so 20 - 4 = 16 cm)

= 2 × 22/7 × 7 × 16

= 2 × 22 × 16

= 704 cm²

Therefore, the area of label is 704 cm².

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