Revision as of 18:06, 4 February 2025

= Problem

you idiot stupidiot

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

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

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 $7^2=\boxed{49\text{ (A)}}$

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-== Problem==
+== 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 <math>14</math>. What is the area of the large square?
+you idiot stupidiot
 <center><asy>pathpen = black+linewidth(0.7);

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

Revision as of 18:06, 4 February 2025

Contents

= Problem

Solution 1

Solution 2

See Also