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

Revision as of 13:47, 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

unitsize(4);
pair B=(0,0);
pair A=(0,6);
pair C=(30,0);
pair D=(30,6);
dot(A);
dot(B):
dot(C);
dot(D);
 (Error making remote request. Unknown error_msg)

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.

Page

Toolbox

Search

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

Revision as of 13:47, 31 December 2019

Problem

Solution

See Also