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

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1992 AHSME Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

See also