A triangle has sides of length $13,30$ and $37$. If the radius of the inscribed circle is $\cfrac{p}{q}$ (where $p$ and $q$ are coprime), then the value of ${q}^{p+3}$ is

Using Heron's formula, the semi-perimeter of the triangle is (13 + 30 + 37)/2 = 40, making the area the square root of (40 - 13)(40 - 30)(40 - 37)(40), which is the square root of 32400, yielding an area of 180. The radius of the inscribed circle is the area divided by the semi-perimeter, so r = 180/40 = 9/2. Here, p = 9 and q = 2, and calculating the value of 2 raised to the power of (9 + 3) gives 2 to the 12th power, which equals 4096.

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