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Difference between revisions of "2012 AMC 10B Problems/Problem 19"

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Revision as of 14:01, 31 December 2019

Problem

In rectangle $ABCD$, $AB=6$, $AD=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended 2 units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of $BFDG$?

$\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}$

Solution

[asy] unitsize(10); pair B=(0,0); pair A=(0,6); pair C=(30,0); pair D=(30,6); pair G=(15,6); pair E=(0,-2); pair F=(15/2,0); dot(A); dot(B); dot(C); dot(D); dot(G); dot(E); dot(F); label("A",(0,6),NW); label("B",(0,0),W); label("C",(30,0),E); label("D",(30,6),NE); label("G",(15,6),N); label("E",(0,-2),SW); label("F",(15/2,0),N); label("15",(A--G),N); label("15",(G--D),N); label("6",(A--B),W); label("2",(B--E),W); label("6",(D--C),W); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(E--D); draw(B--E); [/asy]

Note that the area of $BFDG$ equals the area of $ABCD-\triangle AGB-\triangle DCF$. Since $AG=\frac{AD}{2}=15,$ $\triangle AGB=\frac{15\times 6}{2}=45$. Now, $\triangle AED\sim \triangle BEF$, so $\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5$ and $FC=22.5,$ so $\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}.$

Therefore, \begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*} hence our answer is $\fbox{C}$

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AMC 10 Problems and Solutions

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