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 16  and the triangle has an area of 128 . What are the lengths of sides A and B?

Apr 15, 2016

$8 \setminus \sqrt{5}$ units.

Explanation:

The area of the triangle is $128$. The base is $16$. So the altitude to the base is

$h = 2 \left(\frac{128}{16}\right) = 16$.

In an isosceles triangle the altitude to the base forms two congruent right triangles. In each right triangle:

1) One leg of the right triangle is the altitude, in this case $16$.

2) The other leg is half the base of the isosceles triangle, in this case $8$.

3) The hypoteneuse of the right triangle is one leg of the isosceles triangle. Thus using the Pythagorean theorem:

$\setminus \sqrt{{16}^{2} + {8}^{2}} = \setminus \sqrt{320} = 8 \setminus \sqrt{5}$