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

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Answer 1 · 2016-04-15T23:54:19

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

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

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

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 $16$ $16$ .

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

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

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

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 1616 and the triangle has an area of 128128 . What are the lengths of sides A and B?