# Two corners of an isosceles triangle are at (4 ,9 ) and (9 ,3 ). If the triangle's area is 64 , what are the lengths of the triangle's sides?

Feb 11, 2016

The sides are:
Base, $b = \overline{A B} = 7.8$
Equal sides, $\overline{A C} = \overline{B C} = 16.8$

#### Explanation:

${A}_{\Delta} = \frac{1}{2} b h = 64$
Using the distance formula find b...
$b = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
x_1 = 4; x_2 = 9; y_1 = 9; y_2 = 3
substitute and find h:
$b = \sqrt{25 + 36} = \sqrt{61} \approx 7.81$
$h = 2 \frac{64}{\sqrt{61}} = 16.4$
Now using Pythagoras theorem find the sides, $\overline{A} C$:
$\overline{A} C = \sqrt{\frac{61}{4} + {128}^{2} / 61} = \sqrt{\frac{3 , 721 + 65 , 536}{2}} = 16.8$