Question

Question #5b166

Answer 1

The perimeter is `=12(sqrt2+1)=28.97cm`

Explanation:

Let the length of the hypotenuse be `=12cm`

And one side be `=xcm`

Then by The Pythagoras theorem, the third side is

`=sqrt(144-x^2)`

The perimeter is

`P=x+12+sqrt(144-x^2)`

Deriving with respect to `x`

`(dP)/(dx)=1-x/(sqrt(144-x^2))`

The critical points max. or min) are when `(dP)/(dx)=0`

`1-x/(sqrt(144-x^2))=0`

`x/(sqrt(144-x^2))=1`

`x=sqrt(144-x^2)`

`x^2=144-x^2`

`2x^2=144`

`x^2=72`

`x=6sqrt2`

Therefore,

The max. perimeter is

`P=6sqrt2+12+sqrt(144-72)=6sqrt2+12+6sqrt2`

`=12(sqrt2+1)cm=28.97 cm`

graph{x+12+sqrt(144-x^2) [-28.8, 53.4, -1.12, 40.03]}

Answer 2

The perimeter can be of any value (up to infinity). See explanation

Explanation:

The question is not precise enough and the answer is: The perimeter can be any number up to infinity.

.

As you see in the picture the lengths of side `x` can be any real positive number and the corresponding hypotenuse would be::

`y=sqrt(x^2+12^2)`