# Problem of the Day

In the figure above, square has an area of 81, and points and trisect . If the area of is 24, what is the length of ?

A.

B.

C.

D.

E.

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B. $\displaystyle \frac{42}{16}$

See the Solution

### Solution

[latexpage]

The area of square $ABCE$ is 81, so the lengths of $\overline{AB}$ and $\overline{AE}$ are both equal to 9. Points $F$ and $G$ trisect $\overline{AB}$, so $AF=FG=GB=3$. The area of $\triangle{FGD}$ is 24, with base $FG=3$. Using the formula for the area of a triangle, $\displaystyle \frac{1}{2}\cdot 3\cdot AD=24$ and $AD=16$. If $AD=16$, then $ED=7$. $\triangle{AGD}$ and $\triangle{EHD}$ are similar, so the lengths of corresponding sides are in the same ratio. Using this, we can say $\displaystyle \frac{AG}{AD}=\frac{EH}{ED}$. Substituting known values gives us $\displaystyle \frac{6}{16}=\frac{HE}{7}$, and $\displaystyle HE=\frac{42}{16}$.