Total surface area of a cylinder of height H & base radius r is double that of another cylinder of height h & base radius r. Prove h=(H-r)/2 

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You need to remember that the total surface of cylinder is the sum of two times area of base and the surface area of lateral part such that:

TSA = 2A base + `h*(2pi*r)`

You need to evaluate TSA of cylinder that has the height H and the radius r such that:

`TSA_(H,r) = 2*pi*r^2 + 2H*pi*r`

You need to factor out `2pi*r`  such that:

`TSA_(H,r) = 2pi*r(r + H)`

The problem provides the information that `TSA_(H,r)`  is double of `TSA_(h,r).`

You need to evaluate `TSA_(h,r)`  such that:

`TSA_(h,r) = 2pi*r(r + h)`

If you multiply by 2 both sides yields:

`2TSA_(h,r) = 4pi*r(r + h)`

Since `TSA_(H,r) = 2TSA_(h,r),`  then you may substitute `TSA_(H,r)`  for  `2TSA_(h,r)`  such that:

`2pi*r(r + H) = 4pi*r(r + h)`

You need to divide by `2pi*r`  both sides such that:

`(r + H) = 2(r + h) `

You need to open the brackets such that:

`r + H = 2r + 2h`

You need to isolate 2h to one side such that:

`r + H - 2r = 2h`

`H - r = 2h`

You need to divide by 2 such that:

`h = (H-r)/2`

The last line proves what problem demands, hence `h = (H-r)/2`  under given conditions.

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