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)

(More basic) Solution 2 (bad)

$\overline{DE}$ is midway from $A$ to $\overline{BC}$ , and $DE = \frac{BC}{2}$ . Therefore, $DE$ is a quarter of the area, which is $30$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is $5$ times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$ .

2018 AMC 10A (Problems • Answer Key • Resources)
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 (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

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

Page

Toolbox

Search

2018 AMC 10A Problems/Problem 24

Contents

Problem

Solution

(More basic) Solution 2 (bad)

See Also