# NCERT Solutions for Class 9 Maths Exercise 13.2

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NCERT solutions for Class 9 Maths Surface Areas and Volumes

## NCERT Solutions for Class 9 Mathematics Surface Areas and Volumes

###### 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.

Ans. Given: Height of cylinder = 14 cm, Curved Surface Area = 88 cm2

Let radius of base of right circular cylinder = cm

= 88

1 cm

Diameter of the base of the cylinder = = 2 x 1 = 2 cm

###### 2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?

Ans. Given: Diameter = 140 cm

Radius = 70 cm = 0.7 m

Height of the cylinder = 1 m

Total Surface Area of the cylinder

=

=

= 2 x 22 x 0.1 x 1.7 = 7.48 m2

Hence 7.48 m2 metal sheet is required to make the close cylindrical tank.

###### 3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. [See fig.]. Find its:

(i) Inner curved surface area

(ii) Outer curved surface area

(iii) Total surface area

Ans. (i) Length of the pipe = 77 cm, Inner diameter of cross-section = 4 cm

Inner radius of cross-section = 2 cm

Inner curved surface area of pipe = =

= 2 x 22 x 2 x 11 = 968 cm2

(ii) Length of pipe = 77 cm, Outer diameter of pipe = 4.4 cm

Outer radius of the pipe = 2.2 cm

Outer surface area of the pipe =

=

= 44 x 2.2 x 11 = 1064.8 cm2

(iii) Now there are two circles of radii 2 cm and 2.2 cm at both the ends of the pipe.

Area of two edges of the pipe = 2 (Area of outer circle – area of inner circle)

= =

= =

= = 5.28 cm2

Total surface area of pipe

= Inner curved surface area + Outer curved surface area + Area of two edges

= 968 + 1064.8 + 5.28 = 2038.08 cm2

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.

Ans. Diameter of roller = 84 cm

Radius of the roller = 42 cm

Length (Height) of the roller = 120 cm

Curved surface area of the roller =

= = 31680 cm2

= 3.1680 m2

Now area leveled by roller in one revolution = 3.1680 m2

Area leveled by roller in 500 revolutions

= 3.1680 x 500 = 1584.0000 = 1584 m2

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of white washing the curved surface of the pillar at the rate of Rs. 12.50 per m2.

Ans. Diameter of pillar = 50 cm

Radius of pillar = 25 cm = m

Height of the pillar = 3.5 m

Now, Curved surface area of the pillar

=

= = m2

Cost of white washing 1 m2 = Rs. 12.50

Cost of white washing m2

= 12.50 x = Rs. 68.75

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.

Ans. Curved surface area of the cylinder

= 4.4 m2, Radius of cylinder = 0.7 m

Let height of the cylinder =

= 4.4

= 1 m

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:

(i) its inner curved surface area.

(ii) thecosr of plastering this curved surface at the rate of Rs. 40 per m2.

Ans. Inner diameter of circular well = 3.5 m

= = 1.75 m

And Depth of the well = 10 m

(i) Inner surface area of the well =

= = 110 m2

(ii) Cost of plastering 1 m2 = Rs. 40

Cost of plastering 100 m2 = 40 x 110 = Rs. 4400

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 8. Find:

(i) the lateral or curved surface area of a petrol storage tank that is 4.2 m in diameter and 4.5 m high.

(ii) how much steel was actually used if of the steel actually used was wasted in making the tank?

Ans. (i) Diameter of cylindrical petrol tank = 4.2 m

Radius of the cylindrical petrol tank = = 2.1 m

And Height of the tank = 4.5 m

Total surface area of the cylindrical tank = $2\pi rh (r+h)$

= $2 \times {{22} \over 7} \times 2.1 ( 2.1 + 4.5)$ = 87.12 m2

(ii) Let the actual area of steel used be meters

Since of the actual steel used was wasted, the area of steel which has gone into the tank.

Then, steel actually used = =  of its Total Surface area

= $2\pi r\left( {r + h} \right)$

$\Rightarrow$          ${{11} \over {12}}x = 2 \times {{22} \over 7} \times 2.1\left( {2.1 + 4.5} \right)$

$\Rightarrow$          ${{11} \over {12}}x = 44 \times 0.3 \times 6.6$

$\Rightarrow$          ${{11} \over {12}}x = 87.12$

$x = 87.12 \times {{12} \over {11}}$ = 95.04 m2

Hence steel actually used is 95.04 m2.

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 9. In the adjoining figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. [See fig.]

Ans. Height of each of the folding at the top and bottom = 2.5 cm

Height of the frame (H) = 30 cm

Diameter = 20 cm

Now cloth required for covering the lampshade

= CSA of top part + CSA of middle part + CSA of bottom part

=

=

=

=

= 2200 cm2

NCERT Solutions for Class 9 Maths Exercise 13.2

###### 10. The students of a Vidyalayawere asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Ans. Radius of a cylindrical pen holder = 3 cm

Height of the cylindrical pen holder

= 10.5 cm

Cardboard required for pen holder = CSA of pen holder + Area of circular base

= =

= = 226.28 cm2

Since Cardboard required for making 1 pen holder = 226.28 cm2

Cardboard required for making 35 pen holders = 226.28 x 35 = 7919.8 cm2

= 7920 cm2 (approx.)

## NCERT Solutions for Class 9 Maths Exercise 13.2

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