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 $FDBG?$

$60$

$65$

$70$

$75$

$80$

Answer: D

Concepts:angle bisector theorem midpoint trapezoid area ratio

Difficulty rating: 2040

Solution:

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

By the Angle Bisector Theorem, $BG:GC=AB:AC=5:1,$ so $BG=\dfrac{5a}{6}.$ Because $DE$ is a midsegment, $DE=\dfrac a2.$ Applying the same theorem in $\triangle ADE$ gives $DF:FE=AD:AE=5:1,$ so $DF=\dfrac56DE=\dfrac{5a}{12}.$

The trapezoid's height is $h/2,$ and $ah/2=120.$ Its average base length is $\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.$ Therefore, its area is $\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.$ Thus, D is the correct answer.

2018 AMC 10A Problem 24

Answer: D

Difficulty rating: 2040

Solution:

Problem 24 in Other Years