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

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==Solution==
 
==Solution==
  
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The easiest way to find the area would be to find the area of <math>ABCD</math> and subtract the areas of <math>ABG</math> and <math>CDF.</math> You can easily get the area of <math>ABG</math> because you know <math>AB=6</math> and <math>AG=15</math>, so <math>ABG</math>'s area is <math>15\cdot 6/2=45</math>. However, for triangle <math>CDF,</math> you don't know <math>CF.</math> However, you can note that triangle <math>BEF</math> is similar to triangle <math>CDF</math> through AA. You see that <math>BE/DC=1/3.</math> So, You can do <math>BF+3BF=30</math> for <math>BF=15/2,</math> and <math>CF=3BF=3(15/2)=45/2.</math> Now, you can find the area of <math>CDF,</math> which is <math>135/2.</math> Now, you do <math>[ABCD]-[ABG]-[CDF]=180-45-135/2=135-135/2=\boxed{135/2},</math>
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Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>.  
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Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5</math> So <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math> 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).
 
which is answer choice (C).
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== See Also ==
 
== See Also ==
  

Revision as of 07:25, 9 March 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*} which is answer choice (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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