Difference between revisions of "1995 AJHSME Problems/Problem 18"

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<math>\text{(A)}\ 25 \qquad \text{(B)}\ 44 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 75</math>
 
<math>\text{(A)}\ 25 \qquad \text{(B)}\ 44 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 75</math>
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==Solution==
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The area taken up by the L's is <math>4*\frac{3}{16}=\frac{3}{4}</math> of the area of the whole square. What remains has <math>\frac{1}{4}</math> of the area of the larger square. <math>\frac{100*100}{4}=\frac{100}{2}*\frac{100}{2}=50*50</math> is the area of the smaller square, so its side length is 50. <math>\text{(C)}</math>

Revision as of 14:03, 5 July 2012

Problem

The area of each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square?

[asy] draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle); draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle); draw((0,1)--(0,2)); draw((1,0)--(2,0)); draw((0,-1)--(0,-2)); draw((-1,0)--(-2,0)); [/asy]

$\text{(A)}\ 25 \qquad \text{(B)}\ 44 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 75$

Solution

The area taken up by the L's is $4*\frac{3}{16}=\frac{3}{4}$ of the area of the whole square. What remains has $\frac{1}{4}$ of the area of the larger square. $\frac{100*100}{4}=\frac{100}{2}*\frac{100}{2}=50*50$ is the area of the smaller square, so its side length is 50. $\text{(C)}$