# 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 32  and the triangle has an area of 16 . What are the lengths of sides A and B?

Nov 13, 2017

$\sqrt{257} \approx 16.03$

#### Explanation:

First things first - we start off with a diagram!

Not the most elegant diagram in the world, but it does the job. By convention, the side opposite angle $A$ is $a$, opposite angle $B$ is $b$ and $C$ is $c$. $h$ is the perpendicular bisector of AB at M - it cuts AB in half at a right angle.

Let $a = b = x$. Our job is to find $x$.

$A r e {a}_{\triangle} = \frac{1}{2} b h$
$16 = \frac{1}{2} \times 32 h$
$16 = 16 h$
$h = 1$

Now, we can apply Pythagoras' Theorem in one of the right-angled triangles. We have the height $h = 1$ and the base$= 16$, so we can work out x.

${1}^{2} + {16}^{2} = {x}^{2}$
$1 + 256 = {x}^{2}$
${x}^{2} = 257$
$x = \sqrt{257} \approx 16.03$