The lateral surface area of a closed cylindrical petrol storage tank that is 4.2m in diameter and 4.5m high.How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.
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Hello!
✓ Radius, r = d / 2 = 2. 1 m
✓ Height of the cylinder = 4.5 m
Then,
♦ Lateral surface area of cylinder :
= 2πrh
= 2 x 22 / 7 × 2. 1 × 4.5 = 59.4 m²
♦ Total surface area of cylinder :
= 2πr (r + h)
= 2 x 22 / 7 x 2.1 (2.1 + 4.5)
= 2 x 22 / 7 x 2.1 x (6.6)
= 87 .12 m²
Let the amount of steel wasted be = x
∴ Steel wasted = x / 12
Amount of steel required - Amount of steel wasted = TSA
⇒ x - x / 12 = 87. 12
⇒ 11 / 12 x = 87. 12
⇒ x = 87. 12 × 12 / 11 ⇒ x = 95.04 m²
Hence,
95.04 m² sheet was actually used for making the tank.
Cheers!
✓ Radius, r = d / 2 = 2. 1 m
✓ Height of the cylinder = 4.5 m
Then,
♦ Lateral surface area of cylinder :
= 2πrh
= 2 x 22 / 7 × 2. 1 × 4.5 = 59.4 m²
♦ Total surface area of cylinder :
= 2πr (r + h)
= 2 x 22 / 7 x 2.1 (2.1 + 4.5)
= 2 x 22 / 7 x 2.1 x (6.6)
= 87 .12 m²
Let the amount of steel wasted be = x
∴ Steel wasted = x / 12
Amount of steel required - Amount of steel wasted = TSA
⇒ x - x / 12 = 87. 12
⇒ 11 / 12 x = 87. 12
⇒ x = 87. 12 × 12 / 11 ⇒ x = 95.04 m²
Hence,
95.04 m² sheet was actually used for making the tank.
Cheers!
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