Exercise 11.1 Exercise 11.2 Exercise 11.3 Exercise 11.4

# 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?

Sol :

We know that,

Total surface area of the cuboid = 2 (*lh* +* bh* +* lb*)

Total surface area of the cube = 6 (*l*)^{2}

Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm^{2}

= [2(2400 + 2000 + 3000)] cm^{2}

= (2 × 7400) cm^{2}

= 14800 cm^{2}

Total surface area of cube (b) = 6 (50 cm)^{2} = 15000 cm^{2}

Thus, the cuboidal box (a) will require lesser amount of material.

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?

Sol :

Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80)]

= 2[3840 + 1152 + 1920]

= 13824 cm^{2}

Total surface area of 100 suitcases = (13824 × 100) cm^{2} = 1382400 cm^{2}

Required tarpaulin = Length × Breadth

1382400 cm^{2} = Length × 96 cm

Length = = 14400 cm = 144 m

Thus, 144 m of tarpaulin is required to cover 100 suitcases.

Question 3

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

Sol :

Given that, surface area of cube = 600 cm^{2}

Let the length of each side of cube be *l*.

Surface area of cube = 6 (Side)^{2}

600 cm^{2} = 6*l*^{2}

*l*^{2}= 100 cm^{2}

*l* = 10 cm

Thus, the side of the 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?

Sol :

Length (*l*) of the cabinet = 2 m

Breadth (*b*) of the cabinet = 1 m

Height (*h*) of the cabinet = 1.5 m

Area of the cabinet that was painted = 2*h* (*l* + *b*) + *lb*

= [2 × 1.5 × (2 + 1) + (2) (1)] m^{2}

= [3(3) + 2] m^{2}

= (9 + 2) m^{2}

= 11 m^{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 area is painted. How many cans of paint will she need to paint the room?

Sol :

Given that,

Length (*l*) = 15 m, breadth (*b*) = 10 m, height (*h*) = 7 m

Area of the hall to be painted = Area of the wall + Area of the ceiling

= 2*h* (*l* + *b*) + *lb*

= [2(7) (15 + 10) + 15 ×10] m^{2}

= [14(25) + 150] m^{2}

= 500 m^{2}

It is given that 100 m^{2} area can be painted from each can.

Number of cans required to paint an area of 500 m^{2}

=

Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Question 6

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

Sol :

Describe how the two figures at the right are alike and how they are different. Which box 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 much sheet of metal is required?

Sol :

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

m^{2}

= 440 m^{2}

Thus, 440 m^{2} sheet of metal is required.

Question 8

The lateral surface area of a hollow cylinder is 4224 cm^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Sol :

A hollow cylinder is cut along its height to form a rectangular sheet.

Area of cylinder = Area of rectangular sheet

4224 cm^{2} = 33 cm × Length

Thus, the length of the rectangular sheet is 128 cm.

Perimeter of the rectangular sheet = 2 (Length + Width)

= [2 (128 + 33)] cm

= (2 × 161) cm

= 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.

Sol :

In one revolution, the roller will cover an area equal to its lateral surface area.

Thus, in 1 revolution, area of the road covered = 2π*rh*

In 750 revolutions, area of the road covered

=

= 1980 m^{2}

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.

Sol :

Height of the label = 20 cm − 2 cm − 2 cm = 16 cm

Radius of the label

Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm.

Area of the label = 2π (Radius) (Height)