1989 AMC 12 Problem 3

Attempt Problem 3 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

24

36

64

81

96

Equal perimeters force the three rectangle widths to be equal

Express one rectangle’s perimeter using the square’s side length

Let the square have side

s.

Since every rectangle has length

s

and the same perimeter, their three widths are equal and are each

\frac{s}{3}.

Thus

2\left(s+\frac{s}{3}\right)=24,

s=9

and the square’s area is

81.

Thus the correct answer is D.

Problem 3 in other years