Question

Yo!
Your Question :o

Find:

(1.) 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 . [EASY]

(2.)How much steel was actually used , if 1/12 of the steel actually used was wasted in making the tank .

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Answer 1

Height (h) of cylindrical tank = 4.5 m

Radius (r) of the circular end of cylindrical tank =

(i) Lateral or curved surface area of tank = 2πrh



= (44 × 0.3 × 4.5) m2

= 59.4 m2

Therefore, CSA of tank is 59.4 m2.

(ii) Total surface area of tank = 2πr (r + h)



= (44 × 0.3 × 6.6) m2

= 87.12 m2

Let A m2 steel sheet be actually used in making the tank.



Therefore, 95.04 m2 steel was used in actual while making such a tank.

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Answer 2

Hola!

Here is your answer -

Given - 1] Diameter of Cylindrical tank - 4.2 m

Height = 4.5m

2] 1/2 steel used was wasted in making the tank.

To find -

1] Lateral Surface area/ Curved surface area of tank.

2] How much steel was used, if 1/2 was wasted in making the tank.

Calculation -

Radius if tank = d/2 = 4.2/2

=> 2.1m

Height = 4.5m

C.S.A of cylinder =

2πrh

= 2 × 22/7 × 2.1 × 4.5

= 2 × 22 × 0.3 × 4.5

= 44 × 0.3 × 4.5

= 13.2 × 4.5

= 59.4 m²

Answer - C.S.A of tank = 59.4m²

2]

Amount of steel used - Amount of steel wasted will be the T.S.A of the cylindrical tank

Therefore we will find the total surface of tank first -

T.S.A of cylinder -

2πr (r + h)

= 2 × 22/7 × 2.1 ( 2.1 + 4.5)

= 2 × 22 × 0.3 ( 6.6)

= 44 × 1.98

= 87. 12 m²

Now,

Let the amount of steel used in making the tank be x

Therefore,

Steel wasted will be => x/12

Amount of steel used - Steel wasted = T.S.A

so,

x - x/12 = 87.12

11/12x = 87.2

therefore,

x = 87.2 × 12/11

x = 95.04m²

Therefore,

the amount of steel used for making the tank is

95.04 m²

Hope it helps!