In a right triangle, the longer leg and the hypotenuse have consecutive integer lengths whose sum is $121$ inches. In inches, what is the perimeter of the triangle?
We can use the fact that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse:
Let the longer leg be x+1, and let the hypotenuse be x+2. Then we can write:
$$(x+1)^2 + x^2 = (x+121)^2$$
Expanding and simplifying:
$$2x^2 + 2x - 360 = 0$$
Using the quadratic formula:
$$x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-360)}}{2(2)} = \frac{-1 \pm \sqrt{7}}{2}$$
Since the length of a side of a triangle must be positive, we take the positive solution:
$$x = \frac{-1 + 12 \sqrt{7}}{2}$$
Therefore, the longer leg has length $\frac{-1 + 12 \sqrt{7}}{2} + 1 = \frac{1 + 12 \sqrt{7}}{2}$, and the hypotenuse has length $\frac{-1 + 12 \sqrt{7}}{2} + 2 = \frac{3 + 12 \sqrt{7}}{2}$. Their sum is:
$$\frac{1 + 12 \sqrt{7}}{2} + \frac{3 + 12 \sqrt{7}}{2} = 2 + 24 \sqrt{7}$$
The shorter leg has length $\sqrt{(\frac{1 + 12 \sqrt{7}}{2})^2 + (\frac{-1 + 12 \sqrt{7}}{2})^2} = 2 \sqrt{7}$.
Therefore, the perimeter of the triangle is $2 + 24 \sqrt{7} + 24 \sqrt{7} = 48 \sqrt{7} + 2$ inches.
