what is its meaning...plzz explain me.....9 ka second part..
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The question says that the steel that has been wasted is the 1/12th part of the amount of steel that was actually required to make the tank. By this, we can equate the total surface area of the tank as the difference between the required and wasted steel, that has been explained in the image.
Ask if you have doubt at any step. Hope this helps.
Ask if you have doubt at any step. Hope this helps.
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(i) Height of tank h = 4.5m, radius = 2.1m.
L.S.A = 2pirh
= 2 * 22/7 * 2.1 * 4.5
= 59.4m^2.
TSA = 2pirh + 2pir^2
= 59.4 + 2 * 22/7 * 2.1 * 2.1
= 87.12m^2.
(2) Let the steel used be x.
Given that 1/12 of the steel actually used was wasted = x - 1/12
= 11x/12
11x/12 = 87.12(from (1))
x = 87.12 * 12/11
= 95.04m^2.
The Actual Steel used = 95.04m^2.
Hope this helps!
L.S.A = 2pirh
= 2 * 22/7 * 2.1 * 4.5
= 59.4m^2.
TSA = 2pirh + 2pir^2
= 59.4 + 2 * 22/7 * 2.1 * 2.1
= 87.12m^2.
(2) Let the steel used be x.
Given that 1/12 of the steel actually used was wasted = x - 1/12
= 11x/12
11x/12 = 87.12(from (1))
x = 87.12 * 12/11
= 95.04m^2.
The Actual Steel used = 95.04m^2.
Hope this helps!
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