### Instruction: What is the radius of the circle inscribed in triangle $ABC$ if $AB = 22, AC=12,$ and $BC=14$? Express your answer in simplest radical form.### Response: I know that the radius of the inscribed circle is related to the area of the triangle and the perimeter of the triangle by the formula $r = \frac{A}{s}$, where $A$ is the area and $s$ is the semi-perimeter.
I can find the area of the triangle using Heron's formula, which says that $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $a, b, c$ are the side lengths.
To use Heron's formula, I need to find the semi-perimeter $s$, which is just half of the perimeter.
The perimeter is $22 + 12 + 14 = 48$, so the semi-perimeter is $s = 24$.
Now I can plug in the values of $s, a, b, c$ into Heron's formula and simplify: $A = \sqrt{24(24-22)(24-12)(24-14)} = \sqrt{24(2)(12)(10)} = \sqrt{5760} = \sqrt{144 \cdot 40} = 12\sqrt{40}$.
To simplify the radical further, I can factor out the largest perfect square that divides 40, which is 4.
So $A = 12\sqrt{40} = 12\sqrt{4 \cdot 10} = 12\sqrt{4}\sqrt{10} = 12(2)\sqrt{10} = 24\sqrt{10}$.
Now that I have the area, I can use the formula for the radius and simplify: $r = \frac{A}{s} = \frac{24\sqrt{10}}{24} = \sqrt{10}$.