1990 AHSME Problems/Problem 7
Problem
A triangle with integral sides has perimeter . The area of the triangle is
Solution
The shortest side must be or . However none of form triangles, so the shortest side must be . Then is degenerate, so the sides must be .
This can be cut in half and reassembled into a rectangle with one side and diagonal . By Pythagoras its area is then which means
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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