## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.2

(If values are not given for ‘π’ Assume π = \(\frac{22}{7}\))

Question 1.

The curved surface area of a right circular cylinder of a height 14 cm is 88 cm^{2}. Find the diameter of the base of the cylinder.

Solution:

Lateral surface area of cylinder which has height 14 cm is 88 cm^{2}.

Diameter of bottom of cylinder, d = ?

h = 14 cm, L.S.A. = 88 cm^{2}, d = ?

Lateral surface area of cylinder, L.S.A.

L.S.A.= 2πrh = 88

88 × r = 88

∴ r = \(\frac{88}{88}\) ∴ r = 1 cm.

∴ Diameter, d = 2r =2 × 1 = 2 cm.

Question 2.

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

Solution:

Height of cylindrical water tank, h = 1 m

Diameter of bottom, d = 140 cm.

h = 1 m, d = 140 cm,

r = 0.7 m.

Area of metal sheet required to prepare water tank, A

A = 2πr (r + h)

= 44 × 0.1 × 1.7

= 4.4 × 1.7

A = 7.48 m^{2}.

Question 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. 13.11). Find its

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

Solution:

Length of metal pipe, l = 77 cm,

h = 77 cm.

Inner diameter = 4 cm, ∴ r_{1} = 2 cm.

Outer diameter= 4.4 cm. ∴ r_{2} = 2.2 cm.

(i) Inner surfce area of pipe (C.S.A) = 2πr_{1}h

= 2 × 22 × 2 × 11

= 968 cm^{2}.

(ii) Outer surface area of pipe (C.S.A) = 2πr_{2}h

= 2 × 22 × 2.2 × 11

= 44 × 24.2

= 1064.8 cm^{2}.

(iii) Total Surface area of Pipe (T.S.A.) = Inner CSA + Outer CSA + Area of circular ring.

= 2πr_{1}h + 2πr_{2}h + (2πr_{2}^{2}h – 2πr_{1}^{2}h)

= 2032.8 + 5.28

∴ T.S.A. = 2038.08 cm^{2}.

Question 4.

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

Solution:

Roller is in the form of cylinder.

Diameter, d = 84 cm,

l = h = 120 cm ⇒ 1.2 m.

∴ Curved surface area of Roller = 2 πrh

= 44 × 0.072

= 3.168 m^{2}

∴ Area of 1 round of roller = 3.168 m^{2}.

Area of 500 rounds … ? …

= 3.168 × 500

= 1584 m^{2}.

∴ Total area of a playground is 1584 m^{2}.

Question 5.

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m^{2}.

Solution:

Diameter of cylindrical pole, d = 50 cm.

∴ r = 0.25 m

height of pole, h = 3.5 m.

Curved surface area of pole = 2πrh

= 44 × 0.25 × 0.5

= 5.5 m^{2}.

Cost of painting for 1 m2 is Rs. 12.50,

Cost of painting for 5.5 m^{2} is … ? …

= 5.5 × 12.50

= Rs.68.75.

Question 6.

Curved surface area of a right circular cylinder is 4.4 m^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:

Curved surface arc of cylinder = 4.4 m^{2}

radius, r = 0.7 m height, h = ?

C.S.A. of cylinder, A= 2πrh

2πrh = 4.4

44 × 0.1 × h = 4.4

4.4 h = 4.4

∴ h = 1 m.

Question 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) the cost of plastering this curved surface at the rate of Rs. 40 per m^{2}.

Solution:

Inner diameter of circular well, d = 3.5 m.

Radius, r = \(\frac{3.5}{2}\) m

Depth of the well – height, h = 10 m.

(i) Inner surfacer area of circular well, C.S.A. = 2πrh

C.S.A. = 2πrh

= 22 × 0.5 × 10

= 110 m^{2}

(ii) Cost of plastering of area 1 m^{2} is Rs. 40

Cost of plastering 110 m^{2}… ? …

= 110 × 40

= Rs. 4400.

Question 8.

In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Diameter of cylindrical pipe, d = 5 cm

= 0.5 m

Length of pipe = height, h = 28 m.

Curved surface area of pipe = ?

C.S.A.= 2πrh

Question 9.

Find

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

(ii) how much steel was actually used, if \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.

Solution:

(i) Petrol tank is in the form of cylinder.

Diameter, d = 4.2 m

h = 4.5 m

TSA of petrol tank = 2πr(r + h)

= 44 × 0.3 × 6.6

= 87.12 m^{2}

(ii) ∴ Area of iron required for manufacturing tanker = 87.12 m^{2}.

If iron used is 1 sq. m then there is a loss of \(\frac{1}{12}\).

Iron required except loss = \(1-\frac{1}{12}\)

= \(\frac{11}{12}\) m^{2}

Area of \(\frac{11}{12}\) m^{2} is 87.12 m^{2}.

Area of 1 m^{2} … ? …

= 95.04 m^{2}.

Question 10.

In Fig. 13.12, 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.

Solution:

Diameter of cylindrical frame, d= 20 cm,

∴ r = 10 cm.

height, h = 30 cm.

For folding 2.5 cm is required.

∴ Total height = 30 + 2.5 + 2.5

= 35 cm.

Curved surface area of cloth covered, A

A= 2πrh

= 44 × 50

= 2200 cm^{2}.

Question 11.

The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using card-board. 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?

Solution:

Penholder is in the shape of a cylinder, which one of the edges is open and another is closed.

radius, r = 3 cm, height, h = 10.5 cm.

Total cardboard required by one Competitor: = Curved Surface Area + Area of bottom = 2πrh + πr^{2}

∴ Total cardboard required for 35 competitors is :