Question

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 $FDBG?$

$60$

$65$

$70$

$75$

$80$

Solution:

Let $BC = a, BG = x, GC = y,$ and $h$ be the length of the altitude through $A.$

By the angle bisector theorem, we get that $\dfrac{50}{x} = \dfrac{10}{y},$ where $y = a - x.$ Substituting yields $BG = \frac{5a}{6}.$ We also know that $DF = \frac{5a}{12}$ due to similar triangles.

Note that the height of the trapezoid is $\frac{1}{2}h,$ and $\frac{ah}{2} = 120.$ The area of the trapezoid is $\dfrac{5a}{8} \cdot \dfrac{h}{2} = \dfrac{5}{8} \cdot \dfrac{ah}{2} = 75.$ Thus, D is the correct answer.

2018 AMC 10A Problem 24

Solution:

Problem 24 in Other Years