# 2016 UNCO Math Contest II Problems/Problem 1

## Problem

The sum of the lengths of the three sides of a right triangle is 56. The sum of the squares of the lengths of the three sides of the same right triangle is 1250. What is the area of the triangle?

## Solution

Call the lengths of the legs of the triangle a and b. Call the length of the hypotenuse c. Then, by the Pythagorean Theorem, $$a^2+b^2=c^2$$ From the problem statement, we also know that $$a^2+b^2+c^2=1250$$ By substitution, we have $$2c^2=1250$$ $$c^2=625$$ Since c is positive, $c=25$. We are also given that $$a+b+c=56$$ Substituting, we get $$a+b=31$$ Therefore, $$(a+b)^2=a^2+2ab+b^2=31^2=961$$ We also know that $$a^2+b^2+625=1250$$ $$a^2+b^2=625$$ Combining the two equations, we get that $$(a^2+2ab+b^2)-(a^2+b^2)=961-625$$ $$2ab=336$$ Since the legs of a right triangle are perpendicular, the area of the triangle is $\frac{1}{2}ab$. We can divide both sides of our equation by 4 to get this, which yields $$\frac{1}{2}ab=\fbox{84}$$