# 1996 AHSME Problems/Problem 17

## Problem

In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$? $[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("A", A, dir((5, 3.5)--A)); label("B", B, dir((5, 3.5)--B)); label("C", C, dir((5, 3.5)--C)); label("D", D, dir((5, 3.5)--D)); label("E", E, dir((5, 3.5)--E)); label("F", F, dir((5, 3.5)--F)); label("2", (0,1), dir(0)); label("6", (7.5,0), N);[/asy]$ $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$

## Solution

Since $\angle C = 90^\circ$, each of the three smaller angles is $30^\circ$, and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.

$[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("A", A, dir((5, 3.5)--A)); label("B", B, dir((5, 3.5)--B)); label("C", C, dir((5, 3.5)--C)); label("D", D, dir((5, 3.5)--D)); label("E", E, dir((5, 3.5)--E)); label("F", F, dir((5, 3.5)--F)); label("2", (0,1), W); label("x", (9,3.5), E); label("x-2", (.2,5), W); label("y", (5,7), N); label("6", (7.5,0), S);[/asy]$

Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$

Then $x-2 = 6\sqrt{3} - 2$, and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$.

The area of the square is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$.

Using the approximation $\sqrt{3} \approx 1.7$, we get an area of just under $147.6$, which is closest to answer $\boxed{E}$. (The actual area is actually greater, since $\sqrt{3} > 1.7$).