# 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 6  and the triangle has an area of 64 . What are the lengths of sides A and B?

##### 1 Answer
Jul 26, 2016

$\left\mid A \right\mid = \left\mid B \right\mid = \sqrt{{3}^{2} + {\left(\frac{64}{3}\right)}^{2}} \approx 21.5434$

#### Explanation:

Using $\left\mid C \right\mid = 6$ as the base and $64$ as the area
the height of the triangle (relative to $C$) is
$\textcolor{w h i t e}{\text{XXX}} h = \frac{64}{3}$ (Since $\text{Area} = \frac{1}{2} \cdot b a s e \cdot h$)

Since $\triangle A B C$ is isosceles with $\left\mid A \right\mid = \left\mid B \right\mid$
the height (relative to side $C$) bisects the length of $C$ (innto two segments of $3$ each).

Using the Pythagorean Theorem
$\textcolor{w h i t e}{\text{XXX}} \left\mid A \right\mid = \sqrt{{3}^{2} + {h}^{2}} = \sqrt{{3}^{2} + {\left(\frac{64}{3}\right)}^{2}}$

Using a calculator or computer we can evaluate this as
$\textcolor{w h i t e}{\text{XXX}} \left\mid A \right\mid \approx 21.5434$

(and $\left\mid B \right\mid = \left\mid A \right\mid$)