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 $(3\cdot3)^2=81$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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