# Difference between revisions of "1966 AHSME Problems/Problem 14"

## Problem

The length of rectangle $ABCD$ is 5 inches and its width is 3 inches. Diagonal $AC$ is divided into three equal segments by points $E$ and $F$. The area of triangle $BEF$, expressed in square inches, is: $\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}$

## Solution

Draw the rectangle $ABCD$ with $AB$ = $5$ and $AD$ = $3$. We created our diagonal, $AC$ and use the Pythagorean Theorem to find the length of $AC$, which is $\sqrt34$. Since $EF$ breaks the diagonal into $3$ equal parts, the lenght of $EF$ is $\frac {\sqrt34}{3}$. The only other thing we need is the height of $BEF$. Realize that the height of $BEF$ is also the height of right triangle $ABC$ using $AC$ as the base. The area of $ABC$ is $\frac{15}{2}$ (using the side lengths of the rectangle). The height of $ABC$ to base $AC$ is $\frac{15}{2}$ divided by $\sqrt34 \cdot 2$ (remember, we multiply by $2$ because we are finding the height from the area of a triangle which is $\frac{bh}{2}$). That simplfies to $\frac{15}{\sqrt34}$ which equal to $\frac{15 \cdot \sqrt34}{34}$. Now doing all the arithmetic, $\frac {\sqrt34}{3} \cdot \frac{15 \cdot \sqrt34}{34}$ = $\frac {5 \cdot 3}{3 \cdot 2}$ = $\frac {5}{2}$.

## Solution 2

Draw the rectangle $ABCD$ with $AB$ = $5$ and $AD$ = $3$. Next draw a height from vertex $B$ to diagonal $AC$ intersecting at $G$ and similarly from $D$ to $AC$ intersecting at $J$. Let's call this height $h$ and we can notice that $BG$ = $DJ$ = $h$. Since the points $E$ and $F$ are trisecting diagonal $AC$ the bases and heights of triangles $BEA$, $BEF$, $BEC$, $DEA$, $DEF$, and $DEC$ have the same areas. Hence the area of one of these triangles such as $BEF$ is $\frac{5 \cdot 3}{2}$ = $\frac {5}{2}$. So our answer is $\fbox{C}$.

~Math_Genius_164

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