# 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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### 2 Answers

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.**

lucky092569,

the total surface area of a cylinder of hieght H & radius r is

2pie * r^2 + 2pie * r * H

The total surface area of a cylinder of height h & radius r is

2pie * r^2 + 2pie * r * h

Since the total surface area of the first cylinder is double that of the second cylinder,

2pie * r^2 + 2pie * r * H = 2*(2pie * r^2 + 2pie * r * h)

so, 2pie * r^2 + 2pie * r * H = 4pie * r^2 + 4pie * r * h

If we simplify this equation,

-2pie * r^2 = 4pie * r * h - 2pie * r * H

so, -2pie * r^2 = (2pie * r)*(2h - H)

Since the common factor of this equation on both sides of the equation is (2pie * r), we can eliminate this factor from both sides.

-r = 2h - H

If we solve this equation to h,

2h = H - r

hence, h = (H - r)/2