Two corners of an isosceles triangle are at (2 ,4 ) and (3 ,8 ). If the triangle's area is 18 , what are the lengths of the triangle's sides?

Feb 8, 2016

First find the length of the base, then solve for the height using the area of 18.

Explanation:

Using the distance formula ...

length of base $= \sqrt{{\left(3 - 2\right)}^{2} + {\left(8 - 4\right)}^{2}} = \sqrt{17}$

Next, find the height ...

Triangle Area = $\left(\frac{1}{2}\right) \times \left(\text{base")xx("height}\right)$

$18 = \left(\frac{1}{2}\right) \times \sqrt{17} \times \left(\text{height}\right)$

height $= \frac{36}{\sqrt{17}}$

Finally, use Pythagorean theorem to find the length of the two equal sides ...

${\left(h e i g h t\right)}^{2} + {\left[\left(\frac{1}{2}\right) \left(b a s e\right)\right]}^{2} = {\left(s i \mathrm{de}\right)}^{2}$

${\left(\frac{36}{\sqrt{17}}\right)}^{2} + {\left[\left(\frac{1}{2}\right) \left(\sqrt{17}\right)\right]}^{2} = {\left(s i \mathrm{de}\right)}^{2}$

Sides $= \sqrt{\frac{5473}{68}} \approx 8.97$

In summary, the isosceles triangle has two equal sides of length $\approx 8.97$ and a base length of $\sqrt{17}$

Hope that helped