1995 AHSME Problems/Problem 22

Problem

A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths 13, 19, 20, 25 and 31, although this is not necessarily their order around the pentagon. The area of the pentagon is

$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$

Solution

Since the pentagon is cut from a rectangle, the cut-off triangle is right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple. Assigning the shortest side, 13, to be the hypotenuse of this triangle, we see that one possible triple is 5-12-13. Indeed this works, by placing the 31 side opposite from the 19 side and the 25 side opposite from the 20 side, leaving the cutaway side to be, as before, 13.

Following from this, we subtract the area of the triangle from that of the big rectangle: $31\cdot25-\frac{12\cdot5}{2}=775-30=745\Rightarrow \mathrm{(E)}$

Page

Toolbox

Search

1995 AHSME Problems/Problem 22

Problem

Solution