Problem:
A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?
Answer Choices:
A. 457
B. 459
C. 461
D. 463
E. 465
Solution:
Let x be the length of the hypotenuse, and let y and z be the lengths of the legs. The given conditions imply that
y2+z2=x2,y+z=32−x, and yz=40
Thus
(32−x)2=(y+z)2=y2+z2+2yz=x2+80
from which 1024−64x=80, and x=(B)459.
Note: Solving the system of equations yields leg lengths of
81(69+2201) and 81(69−2201)
so a triangle satisfying the given conditions does in fact exist.
The problems on this page are the property of the MAA's American Mathematics Competitions