Question

In rectangle $ABCD,$ $AB=1,$ $BC=2,$ and points $E,$ $F,$ and $G$ are midpoints of $\overline{BC},$ $\overline{CD},$ and $\overline{AD},$ respectively. Point $H$ is the midpoint of $\overline{GE}.$ What is the area of the shaded region?

$\dfrac1{12}$

$\dfrac{\sqrt3}{18}$

$\dfrac{\sqrt2}{12}$

$\dfrac{\sqrt3}{12}$

$\dfrac16$

Solution:

We can find the area of the shaded region by finding the area of \triangleriangle DHC and subtracting out the two unshaded triangles.

Extend $\overline{DH}$ so that it hits $B.$ Let the intersection of $\overline{DB}$ and $\overline{AF}$ be $X.$

We have that \triangleriangle DXF \sim \triangleriangle BXA. Since $AB = 2 \cdot DF,$ we have that $BX = 2 \cdot AX.$

This means that $DX = \dfrac{1}{3} \cdot DB,$ which means that the altitude of \triangleriangle DXF is $\dfrac{1}{3}$ the height of the rectangle.

The area of \triangleriangle DXF is then $\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{6}.$

The area of both unshaded triangles is then $2 \cdot \dfrac{1}{6} = \dfrac{1}{3}.$ The area of \triangleriangle DHC is $\dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2}.$

The area of the shaded region is then $\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}.$

Thus, E is the correct answer.

2014 AMC 10A Problem 16

Solution:

Problem 16 in Other Years