# An isosceles triangle has sides that are sqrt125, sqrt125, and 10 units. What is its area?

Nov 21, 2015

$50 {\text{ units}}^{2}$

#### Explanation:

the area of a triangle is found through $A = \frac{1}{2} b h$. We have the base, but not the length of the height.

Imagine that an altitude is drawn down the center of the triangle, perpendicularly bisecting the base and bisecting the vertex angle. This creates two congruent right triangles inside the original isosceles triangle.

You know each of these triangles has a hypotenuse of $\sqrt{125}$ and a base of $5$, since they're half the original base.

We can use the Pythagorean theorem to figure out the length of the missing side, which is the height of the triangle.

${\left(5\right)}^{2} + {\left(h\right)}^{2} = {\left(\sqrt{125}\right)}^{2}$

$25 + {h}^{2} = 125$

${h}^{2} = 100$

$h = 10$

So, we now know that the base is $10$ units and the height is $10$ units.

Thus, $A = \frac{1}{2} \left(10\right) \left(10\right) = 50 {\text{ units}}^{2}$