Difference between revisions of "1998 AHSME Problems/Problem 10"

Revision as of 14:57, 20 June 2011

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
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

@@ Line 9: / Line 9: @@
 ==Solution==
-{{solution}}
+Let the length of the longer side be <math>x</math>, and the length of the shorter side be <math>y</math>.  We are given that <math>2x+2y=14\implies x+y=7</math>.  However, note that <math>x+y</math> is also the length of a side of the larger square.  Thus the area of the larger square is <math>(x+y)^2=7^2=\boxed{49\text{ (B)}}</math>.
 ==See Also==
 {{AHSME box|year=1998|num-b=9|num-a=11}}

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Difference between revisions of "1998 AHSME Problems/Problem 10"

Revision as of 14:57, 20 June 2011

Problem 10

Solution

See Also