# Two corners of an isosceles triangle are at (2 ,3 ) and (1 ,4 ). If the triangle's area is 64 , what are the lengths of the triangle's sides?

Oct 21, 2016

The 3 sides are $90.5 , 90.5 , \mathmr{and} \sqrt{2}$

#### Explanation:

Let b = the length of the base from $\left(2 , 3\right)$ to $\left(1 , 4\right)$

$b = \sqrt{{\left(1 - 2\right)}^{2} + {\left(4 - 3\right)}^{2}}$

$b = \sqrt{2}$

This cannot be one of the equal sides, because the maximum area of such a triangle would occur, when it is equilateral, and specifically:

$A = \frac{\sqrt{3}}{2}$

This conflicts with our given area, $64 u n i t {s}^{2}$

We can use the Area to find the height of the triangle:

$A r e a = \left(\frac{1}{2}\right) b h$

$64 = \frac{1}{2} \sqrt{2} h$

$h = 64 \sqrt{2}$

The height forms a right triangle and bisects the base, therefore, we can use the Pythagorean theorem to find the hypotenuse:

${c}^{2} = {\left(\frac{\sqrt{2}}{2}\right)}^{2} + {\left(64 \sqrt{2}\right)}^{2}$

${c}^{2} = 8192.25$

$c \approx 90.5$