The total surface area of a hollow cylinder which is open from both sides is 4620 sq.cm,area of base ring is 115.5sq.cm and height 7cm. Find the thickness of the cylinder.
Let R be the external radius of cylinder and r be the internal radius of cylinder.
Then = 2*pie*R*h + 2*pie*r*h + 2 * pie ( R2 - r2) = 4620
pie ( R2 - r2) = 115.5
By substituting pie * ( R2 - r2 ) = 115.5 in first equation , we will get
2*pie*R*h + 2*pie*r*h + 2* 115.5 = 4620
2*pie*R*h + 2*pie*r*h + 231 = 4620
2*pie*R*h + 2*pie*r*h = 4620 - 231 = 4389
Now taking 2 * pie * h as common we get
2 * pie * h ( R + r) = 4389
2 * (22 / 7) * 7 ( R + r) = 4389
44(R + r) = 4389
(R + r ) = 4389 / 44 by cutting it by eleven we get 399 / 4
We nknow that
pie * (R2 - r2) = 115.5 which implies
(22 / 7) * (R - r) ( R + r) = 115.5
R + r = 399 / 4
(22 / 7) * (399 / 4) * (R - r) = 115.5
R - r = (115.5 * 7 * 4) / ( 22 * 399)
R - r = 7 / 9 cm
See R - r is always the thickness please remember.
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