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

By noting the following: \begin{align*}|EFHG| &= |\triangle EGH| + |\triangle EFH| \\ &= \frac{1}{2}|AEHD|+\frac{1}{2}|EHCB| \\ &= \frac{1}{2}[|AEHD|+|EHCB|] \\ &= \frac{1}{2}|ABCD| \\ &= 5 \end{align*} we see that the answer is $\fbox{C}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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