Question

An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of `4` and the triangle has an area of `128`. What are the lengths of sides A and B?

Answer 1

`A=B=10sqrt41`

Explanation:

When in doubt, draw a diagram.

`h` is the perpendicular bisector of `AB` from `C`, and meets `AB` at point `D`.

Sidenote: I have labeled this diagram according to the convention, with angles `A, B` and `C` opposite sides `a, b` and `c` respectively. So where you have `A` and `B`, I have `a` and `b`. Just a note.

We know that `Area=1/2xxbasexxheight`. In here, the base is length 4, and the height is `h`.

`128=1/2xx4h`
`128=2h`
`h=64`

Now, focus just on `triangleBCD` (or `triangle ACD`). To work out the remaining length `a` (and `b=a` since the `triangle` is isosceles) we need to use the Pythagorean Theorem. We have a base of 2, since it is half of the length 4 (remember that `h` is a bisector).

`2^2+64^2=a^2`
`a^2=4100`
`a=sqrt4100=10sqrt41`

Convert back to the format given in the question:

`A=B=10sqrt41`