Isosceles triangle from the inradius and circumradius

Text

An isosceles triangle has inradius $\dfrac32$ and circumradius $\dfrac{25}{8}$. Moreover the base is $6$ long. Determine the slant side and the area of the triangle, and draw it.

Solution

Let $b=6$ be the base, $L$ the slant side, $h$ the height (falling at the midpoint of the base). Half base $=3$, so $L=\sqrt{h^2+9}$ and the area is $\mathcal A=3h$. Circumradius: $R=\dfrac{L^2}{2h}=\dfrac{25}{8}$. Inradius: $r=\dfrac{\mathcal A}{p}=\dfrac{3h}{3+L}=\dfrac32\ \Longrightarrow\ L=2h-3$. Substituting into $L^2=h^2+9$: $(2h-3)^2=h^2+9\ \Longrightarrow\ 3h^2-12h=0\ \Longrightarrow\ h=4.$ Then $L=5$ (check: $L^2=25=16+9$ ✓, and $R=\dfrac{25}{8}$ ✓). The area is $\mathcal A=12$. $\boxed{L=5,\qquad \mathcal A=12.}$