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

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==Problem==
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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>?
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<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>
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[[Category: Introductory Geometry Problems]]
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==Solution==
 
==Solution==
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<asy>
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unitsize(10);
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pair B=(0,0);
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pair A=(0,6);
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pair C=(30,0);
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pair D=(30,6);
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pair G=(15,6);
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pair E=(0,-2);
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pair F=(15/2,0);
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dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(G);
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dot(E);
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dot(F);
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label("A",(0,6),NW);
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label("B",(0,0),W);
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label("C",(30,0),E);
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label("D",(30,6),NE);
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label("G",(15,6),N);
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label("E",(0,-2),SW);
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label("F",(15/2,0),N);
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label("15",(A--G),N);
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label("15",(G--D),N);
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label("6",(A--B),W);
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label("2",(B--E),W);
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label("6",(D--C),E);
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draw(A--B);
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draw(B--C);
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draw(C--D);
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draw(D--A);
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draw(E--D);
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draw(B--E);
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draw(B--G);
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</asy>
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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>
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Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath>
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hence our answer is <math>\fbox{C}</math>
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 +
==Solution 2==
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 +
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>
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==Solution 3 (Coordinate Bash)==
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Let <math>D=(0,0)</math>.
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We know these points from the problem statement:
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<math>C=(6,0), B=(6,30), A=(0,30), G=(0,15), E=(8,30)</math>
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 +
We can use the Shoelace Formula to find the area of quadrilateral <math>BFDG</math>. We know the coordinates of all of the points in <math>BFDG</math> except <math>F</math>. Since <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>, we can use a system of equations to solve for the coordinates of <math>F</math>. The line for <math>\overline{BC}</math> is simply <math>x = 6</math>. The line for <math>\overline{ED}</math> passes through the origin so it has a y-intercept of <math>0</math>, and a slope of <math>\frac{30-0}{8-0}=\frac{30}{8}=\frac{15}{4}</math>. Therefore, the line for <math>\overline{ED}</math> is <math>y=\frac{15}{4} x</math>. Substituting <math>6</math> for <math>x</math>, we find that <math>y = \frac{45}{2}</math>. Therefore, <math>F=(6, \frac{45}{2})</math>. Applying Shoelace on these points gives us that <math>[BFDG] = \boxed{\frac{135}{2}}</math>.
  
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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=135/2,</math>
 
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which makes the answer (C).
 
 
== See Also ==
 
== See Also ==
  

Latest revision as of 13:23, 23 August 2022

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 $6\cdot \frac{15+\frac{15}{2}}{2}=3\cdot \frac{45}{2}=\boxed{\textbf{(C) }\frac{135}{2}}$

Solution 3 (Coordinate Bash)

Let $D=(0,0)$. We know these points from the problem statement: $C=(6,0), B=(6,30), A=(0,30), G=(0,15), E=(8,30)$

We can use the Shoelace Formula to find the area of quadrilateral $BFDG$. We know the coordinates of all of the points in $BFDG$ except $F$. Since $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$, we can use a system of equations to solve for the coordinates of $F$. The line for $\overline{BC}$ is simply $x = 6$. The line for $\overline{ED}$ passes through the origin so it has a y-intercept of $0$, and a slope of $\frac{30-0}{8-0}=\frac{30}{8}=\frac{15}{4}$. Therefore, the line for $\overline{ED}$ is $y=\frac{15}{4} x$. Substituting $6$ for $x$, we find that $y = \frac{45}{2}$. Therefore, $F=(6, \frac{45}{2})$. Applying Shoelace on these points gives us that $[BFDG] = \boxed{\frac{135}{2}}$.

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

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