Difference between revisions of "2010 AMC 12A Problems/Problem 3"

Revision as of 19:41, 12 February 2010

Problem 3

Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$ ?

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray); draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); draw((25,15)--(25,20)--(50,20)--(50,15)--cycle); label("$A$",(0,20),W); label("$B$",(50,20),E); label("$C$",(50,15),E); label("$D$",(0,15),W); label("$E$",(0,25),NW); label("$F$",(25,25),NE); label("$G$",(25,0),SE); label("$H$",(0,0),SW); [/asy]$

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution

Let $EF = FG = GF = HE = s$ , let $AD = BC = h$ , and let $AB = CD = x$ .

$.2s^2 = hs$

$s = 5h$

$.5hx = hs$

$x = 2s = 10h$

$\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}$

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Difference between revisions of "2010 AMC 12A Problems/Problem 3"

Revision as of 19:41, 12 February 2010

Problem 3

Solution

@@ Line 4: / Line 4: @@
 <center><asy>
 unitsize(1mm);
-defaultpen(linewidth(.8pt));
+defaultpen(linewidth(.8pt)+fontsize(8pt));
 draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
@@ Line 10: / Line 10: @@
 draw((0,15)--(0,20)--(25,20)--(25,15)--cycle);
 draw((25,15)--(25,20)--(50,20)--(50,15)--cycle);
+label("$A$",(0,20),W);
+label("$B$",(50,20),E);
+label("$C$",(50,15),E);
+label("$D$",(0,15),W);
+label("$E$",(0,25),NW);
+label("$F$",(25,25),NE);
+label("$G$",(25,0),SE);
+label("$H$",(0,0),SW);
 </asy></center>
 <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math>