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

Revision as of 21:59, 6 July 2018

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

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 $\fbox{C}$

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

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

@@ Line 11: / Line 11: @@
 Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath>
-which is answer choice (C).
+hence our answer is <math>\fbox{C}</math>
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Difference between revisions of "2012 AMC 10B Problems/Problem 19"

Revision as of 21:59, 6 July 2018

Problem

Solution

See Also