2016 UNCO Math Contest II Problems/Problem 1


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?


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 \[\frac{1}{2}ab=\fbox{84}\]

See also

2016 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions