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, $\frac {\sqrt34}{3} \cdot \frac{15 \cdot \sqrt34}{34}$ = $\frac {5 \cdot 3}{3 \cdot 2}$ = $\frac {5}{2}$ .

$\fbox{C}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1966 AHSME Problems/Problem 14

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