Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "The easiest way to find the area would be to find the area of ABCD and subtract the areas of ABG and CDF. You can easily get the area of ABG because you know AB=6 and AG=15, so A...")
 
(35 intermediate revisions by 10 users not shown)
Line 1: Line 1:
−
The easiest way to find the area would be to find the area of ABCD and subtract the areas of ABG and CDF. You can easily get the area of ABG because you know AB=6 and AG=15, so ABG's area is 5. However, for triangle CDF, you don't know CF. However, you can note that triangle BEF is similar to triangle CDF through AA. You see that BE/DC=1/3. So, You can do BF+3BF=30 for BF=15/2, and CF=15/2. Now, you can find the area of CDF, which is 135/2. Now, you do 180-225/2, which turns out to be 135/2, which makes the answer (C).
+
==Problem==
 +
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of quadrilateral <math>BFDG</math>?
 +
 
 +
<math>\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}</math>
 +
[[Category: Introductory Geometry Problems]]
 +
 
 +
==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),E);
 +
draw(A--B);
 +
draw(B--C);
 +
draw(C--D);
 +
draw(D--A);
 +
draw(E--D);
 +
draw(B--E);
 +
draw(B--G);
 +
</asy>
 +
 
 +
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>.  
 +
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>
 +
hence our answer is <math>\fbox{C}</math>
 +
 
 +
==Solution 2==
 +
 
 +
Notice that <math>BFDG</math> is a trapezoid with height <math>6</math>, so we need to find <math>BF</math>. <math>\triangle BFE\sim \triangle ADE</math>, so <math>4\cdot BF = AD</math>. Since <math>AD = 30</math>, <math>BF = \frac{15}{2}</math>. The area of <math>BFDG</math> is <math>6\cdot \frac{15+\frac{15}{2}}{2}=3\cdot \frac{45}{2}=\boxed{\textbf{(C) }\frac{135}{2}}</math>
 +
 
 +
== See Also ==
 +
 
 +
 
 +
{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}
 +
 
 +
{{MAA Notice}}

Revision as of 15:24, 28 August 2020

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 quadrilateral $BFDG$?

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

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),E); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(E--D); draw(B--E); draw(B--G); [/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 $\fbox{C}$

Solution 2

Notice that $BFDG$ is a trapezoid with height $6$, so we need to find $BF$. $\triangle BFE\sim \triangle ADE$, so $4\cdot BF = AD$. Since $AD = 30$, $BF = \frac{15}{2}$. The area of $BFDG$ is $6\cdot \frac{15+\frac{15}{2}}{2}=3\cdot \frac{45}{2}=\boxed{\textbf{(C) }\frac{135}{2}}$

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_19&oldid=132658"