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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...")
 
 
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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).
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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==
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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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== See Also ==
 +
 
 +
 
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{{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 14:23, 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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