For previous exercise – Surface Area and Volumes – Exercise 13.2 – Class IX

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

Solution:

Given, h = 14 cm ; surface area = 88 cm^{2}

Curved surface area of the cylinder = 2πrh

88 = 2 x ^{22}/_{7} x r x 14

r = ^{88×7}/_{2x22x14} = 1 cm

Hence, base diameter of the cylinder = 2r = 2x 1 = 2 cm

**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?**

Solution:

h = 1 m

d = 140 cm

r = ^{d}/_{2} = ^{140}/_{2} = 70 cm = 0.7 m

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

= 2 x ^{22}/_{7} x 0.7 x (1 + 0.7)

= 2 x ^{22}/_{7} x 0.7 x 1.7

= 7.48 m^{2}

**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:

h = 77 cm

Inner diameter = 4 cm ⇒ r = ^{d}/_{2} = ^{4}/_{2 }= 2 cm

Outer diameter = 4.4 cm ⇒ r = ^{d}/_{2}= ^{4.4}/_{2} = 2.2 cm

(i) inner curved surface area = 2πRh

= 2 x ^{22}/_{7} x 2 x 77

= 968 cm^{2}

(ii) outer curved surface area = 2πrh

= 2 x ^{22}/_{7} x 2.2 x 77

= 1064.8 cm^{2}

(iii) total surface area =inner curved surface area + outer curved surface area + two base rings

= 2πrh + 2π Rh + 2π ( R^{2} – r^{2})

= 968 + 1064.8 + 2x^{22}/_{7}x(2.2^{2} – 2^{2})

= 968 + 1064.8 + 5.28

= 2038. 08 cm^{2}

**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 m**^{2}

Solution:

d = 84 cm ; r = ^{d}/_{2}= ^{84}/_{2} = 42 cm

h = 120 cm

Surface area of the roller = 2πrh = 2 x ^{22}/_{7} x 42 x 120

=31680 cm^{2}

Given, it takes 500 complete revolutions to move once over to level a playground. Then the area of the playground = 31680 x 500 = 15840000 cm^{2} = 1584 m^{2}

**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:

d = 50 cm ; r = ^{50}/_{2} = 25 cm

h = 3.5 m

Curved surface area of the pillar = 2πrh = 2 x ^{22}/_{7} x 25 x 3.5 =550m^{2}

The cost of painting the pillar for 1m^{2} is Rs. 12.50

Then the cost of 550m^{2} at the rate ofRs12.50=12.50*550=Rs. 68.75

**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:

Given, curved surface area of the cylinder 4.4 m^{2} and radius, r = 0.7 m

surface area = 2πrh

4.4 = 2x ^{22}/_{7} x 0.7 x h

h = ^{4.4}/_{4.4 } = 1 m

**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 ; r = 3.5 m and h = 10m

(i) inner curved surface area = 2πrh = 2 x ^{22}/_{7} x 3.5 x 10 = 220 m^{2}

(ii) The cost of plastering of curved surface for 1m^{2} is Rs. 40

Then, the cost of plastering for 220m^{2} = 220 x 40 = Rs. 8800

**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:

h = 28 m and d = 5 cm; r = ^{5}/_{2} = 2.5 cm = 0.025 cm

Total surface area = 2πr (r + h) = 2 x ^{22}/_{7} x 0.025 x(0.025 + 28) = 0.1571 x(28.025) = 4.403 m^{2}

**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 1/12 of the steel actually used was wasted in making the tank.**

Solution:

d = 4.2 m ; r = ^{4.2}/_{2}=2.1 m

h = 4.5 m

(i) Curved surface area of the cylinder = 2πrh = 2 x ^{22}/_{7}x 2.1 x 4.5 =59.4 m^{2}

(ii) Total surface area =2πr(r+h)

= 2 x ^{22}/_{7}x 2.1 (2.1 + 4.5)

= 87.12m^{2}

Lets total steel area of steel used be x m^{2}

Area of steel being wasted = ^{1}/_{12} of x m^{2} = ^{x}/_{12} m^{2}

Area of steel used in the tank = (x – ^{x}/_{12}) m^{2} = ^{11x}/_{12} m^{2}

87.12 = ^{11x}/_{12}

x = ^{87.12×12}/_{11} = 95.04 m^{2}

Hence, 95.04 m^{2} of steel was actually used in tank.

**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:

d = 20 cm; r = ^{d}/_{2} = ^{20}/_{2} = 10 cm

h = 30 cm

Circumference of the base of the frame = 2πr = 2 x ^{22}/_{7} x 10 = 62. 86 cm

Height of the cloth required for covering frame (including margin) = (30+2.5+2.5)cm = 35 cm

Also, breadth of the cloth = circumference of the base of the frame

Area of the cloth required for covering the lampshade = length x breadth = 35 x 20 π cm^{2} = 35 x 20 x ^{22}/_{7} cm^{2} = 2200 cm^{2}

**The students of a Vidyalaya were asked to participate in a competition for making and decorating penholder 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?**

Solution:

Here, r = 3 cm, h = 10.5 cm

The penholder have only one base i.e., these are open at one end.

Total surface area of 1 penholder = 2πrh + πr^{2} = πr (2h + r)

= ^{22}/_{7} × 3 (2 × 10.5 + 3) cm^{2}

= ^{22}/_{7} × 3 × 24 cm^{2}

Total surface area of 35 penholder = ^{22}/_{7} × 3 × 24 × 35 cm2 = 7920 cm^{2}

Hence, 7920 cm^{2} of cardboard is needed.

For next exercise – Surface Area and Volumes – Exercise 13.3 – Class IX

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