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 `8` and the triangle has an area of `40`. What are the lengths of sides A and B?

Answer 1

Side A = Side B = 10.77 units

Explanation:

Let side A=side B= `x`

Then drawing a right-angled triangle, with `x` as your hypotenuse and `4` as the length of your bottom side and the final side be `h`

Using pythagoras theorem,

`x^2` = `h^2+4^2`

`x^2=h^2+16`

`h^2=x^2-16`

`h=+-sqrt(x^2-16)`

But since h is a length, it can only be positive

`h=sqrt(x^2-16)`

Area of triangle =

`1/2times8timessqrt(x^2-16) = 40`

`4sqrt(x^2-16)=40`

`sqrt(x^2-16)=10`

`x^2-16=100`

`x^2=116`

`x=+-10.77`

But since x is a side, then it can only be positive

`x=10.77`

Answer 2

`a = b = 2 sqrt{29}`

Explanation:

Hey, small letters for triangle sides.

Isosceles triangle `a=b`, `c=8`, Area `mathcal{A}=40.`

Let's see if we can do this different ways. First, the way the teacher probably expects:

The foot `F` of the altitude `h` from vertex `C` to side `c` is the midpoint of `AB`. So

`mathcal{A} = 1/2 ch`

`40 = 1/2 (8) h`

`h = 10`

`AF=c/2=4` and we have a right triangle with the altitude:

`(c/2)^2 + h^2 = a^2`

`4^2 + 10^2 = a^2`

`a = b = sqrt{116} = 2 sqrt{29}`

Let's just do it directly from Archimedes' Theorem:

`16 mathcal{A} ^2 = 4a^2 c^2 - (b^2 - c^2-a^2)^2`

The sides can be arbitrarily interchanged; I picked the form that gets some cancelling when `a=b`:

`16 mathcal{A} ^2 = 4a^2 c^2 - c^4`

`a^2 = {c^4 + 16 mathcal{A}^2}/{4c^2}`

`a = \sqrt{ {8^4 + 16(40)^2}/{4(8^2)}} =2 sqrt{29} quad sqrt`

That was easier.