1998 AHSME Problems/Problem 10

Problem 10

A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?

$[asy]pathpen = black+linewidth(0.7); D((0,0)--(7,0)--(7,7)--(0,7)--cycle); D((1,0)--(1,6)); D((0,6)--(6,6)); D((1,1)--(7,1)); D((6,7)--(6,1)); [/asy]$

$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$

Solution

Let the length of the longer side be $x$ , and the length of the shorter side be $y$ . We are given that $2x+2y=14\implies x+y=7$ . However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\boxed{49\text{ (B)}}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1998 AHSME Problems/Problem 10

Problem 10

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