# 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?

Nov 11, 2017

$A = B = 10 \sqrt{41}$

#### Explanation:

When in doubt, draw a diagram. $h$ is the perpendicular bisector of $A B$ from $C$, and meets $A B$ 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 $A r e a = \frac{1}{2} \times b a s e \times h e i g h t$. In here, the base is length 4, and the height is $h$.

$128 = \frac{1}{2} \times 4 h$
$128 = 2 h$
$h = 64$

Now, focus just on $\triangle B C D$ (or $\triangle A C D$). 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).

${2}^{2} + {64}^{2} = {a}^{2}$
${a}^{2} = 4100$
$a = \sqrt{4100} = 10 \sqrt{41}$

Convert back to the format given in the question:

$A = B = 10 \sqrt{41}$