## Inscribed Circle In Isosceles Triangle

#### Problem

A circle is inscribed in an isosceles with the given dimensions.

Find the radius of the circle.

#### Solution

Consider the following diagram.

By dropping a perpendicular from the top of the isosceles triangle to the base and using the Pythagorean Theorem we quickly determine that the height of the triangle is 4.

Therefore the area of the isosceles triangle is $\dfrac{6 \times 4}{2} = 12$.

However, we can split the isosceles triangle into three separate triangles indicated by the red lines in the diagram below. Because the radius always meets a tangent at a right angle the area of each triangle will be the length of the side multiplied by the radius of the circle. So the total area of the isosceles triangle is given by $\dfrac{6r}{2} + 2 \times \dfrac{5r}{2} = 8r = 12 \Rightarrow r = \dfrac{3}{2}$.

If the base length of the isosceles triangle is $b$ and the two legs are $a$ then prove that the radius of the inscribed circle is given by,$$r = \frac{b}{2} \sqrt{\frac{2a-b}{2a+b}}$$