# 1992 AHSME Problems/Problem 24

## Problem

Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$. Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$, respectively, with $AE=BF=AG=2$. Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$. The area of quadrilateral $EFHG$ is $\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$

## Solution 1 $\fbox{C}$ Use vectors. Place an origin at $A$, with $B = p, D = q, C = p + q$. We know that $\|p \times q\|=10$, and also $E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q$, and now we can find the area of $EFHG$ by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).

## Solution 2

We note that $ABFG$ is a parallelogram because $AG = BF = 2$ and $AG \parallel BF$. Using the same reasoning, $GFCD$ is also a parallelogram.

Assume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$. Then, the area of parallelogram $ABFG$ is $AB * x$. The area of triangle $EFG$ is $\frac{AB * x}{2}$, which is half of the area of parallelogram $ABFG$.

Likewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$.

Thus, $[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}$

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