Difference between revisions of "2018 AMC 10A Problems/Problem 24"

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== Problem ==
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Triangle <math>ABC</math> with <math>AB=50</math> and <math>AC=10</math> has area <math>120</math>. Let <math>D</math> be the midpoint of <math>\overline{AB}</math>, and let <math>E</math> be the midpoint of <math>\overline{AC}</math>. The angle bisector of <math>\angle BAC</math> intersects <math>\overline{DE}</math> and <math>\overline{BC}</math> at <math>F</math> and <math>G</math>, respectively. What is the area of quadrilateral <math>FDBG</math>?
 
Triangle <math>ABC</math> with <math>AB=50</math> and <math>AC=10</math> has area <math>120</math>. Let <math>D</math> be the midpoint of <math>\overline{AB}</math>, and let <math>E</math> be the midpoint of <math>\overline{AC}</math>. The angle bisector of <math>\angle BAC</math> intersects <math>\overline{DE}</math> and <math>\overline{BC}</math> at <math>F</math> and <math>G</math>, respectively. What is the area of quadrilateral <math>FDBG</math>?
  
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\textbf{(E) }80 \qquad
 
\textbf{(E) }80 \qquad
 
</math>
 
</math>
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== Solution ==
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By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter)
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==See Also==
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{{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}
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{{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}
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{{MAA Notice}}

Revision as of 14:45, 8 February 2018

Problem

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution

By angle bisector theorem, $BG=\frac{5a}{6}$. By similar triangles, $DF=\frac{5a}{12}$, and the height of this trapezoid is $\frac{h}{2}$, where $h$ is the length of the altitude to $BC$. Then $\frac{ah}{2}=120$ and we wish to compute $\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}$. (trumpeter)

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 12 Problems and Solutions

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