1989 AHSME Problems/Problem 3

Problem

A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is

$[asy] draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(3,9), dashed); draw((6,0)--(6,9), dashed);[/asy]$

$\textrm{(A)}\ 24\qquad\textrm{(B)}\ 36\qquad\textrm{(C)}\ 64\qquad\textrm{(D)}\ 81\qquad\textrm{(E)}\ 96$

Solution

Let $x$ be the width of a rectangle so that the side of the square is $3x$ . Since $2(3x)+2(x)=24$ we have $x=3$ . Thus the area of the square is $(3\cdot3)^2=81$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1989 AHSME Problems/Problem 3

Problem

Solution

See also