2018 AMC 10A Problems/Problem 24

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)

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
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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2018 AMC 10A Problems/Problem 24

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