In the diagram above $\triangle{ABC}$ is an isosceles triangle with$|\overline {\rm AB}| =|\overline {\rm BC}| = |\overline {\rm AC}| - 1 $. If the radius of the inscribed circle is $\dfrac{3}{2}$, find the area of $\triangle{ABC}$.
Inradius r = 3/2 = 1.5
Semiperimeter s = (3x-1)/2
Triangle area A = r (s)
r = {sqrt[x2-((x-1)/2)2] *(x-1)/2 } / (3x-1)/2
x = 3 + √7
I'll let you find the area of ΔABC
