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

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==Solution 2==
 
==Solution 2==
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Expand the small square so it basically equals the area of the large square.  Two of the sides of the rectangles shrink to zero.  The other two sides expand to equal the length of the large outer square, and have a length of <math>\frac{14}{2} = 7</math>.  Thus, the area of the larger square is <math>7^2=\boxed{49\text{ (B)}}</math>
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Expand the small square so it basically equals the area of the large square.  Two of the sides of the rectangles shrink to zero.  The other two sides expand to equal the length of the large outer square, and have a length of <math>\frac{14}{2} = 7</math>.  Thus, the area of the larger square is <math>7^2=\boxed{49\text{ (A)}}</math>
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1998|num-b=9|num-a=11}}
 
{{AHSME box|year=1998|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 13:29, 5 July 2013

Problem

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 1

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 $(x+y)^2=7^2=\boxed{49\text{ (A)}}$.

Solution 2

Expand the small square so it basically equals the area of the large square. Two of the sides of the rectangles shrink to zero. The other two sides expand to equal the length of the large outer square, and have a length of $\frac{14}{2} = 7$. Thus, the area of the larger square is $7^2=\boxed{49\text{ (A)}}$

See Also

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

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