# 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 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...

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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