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Difference between revisions of "1992 AHSME Problems/Problem 24"

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== Solution 2 ==
 
== Solution 2 ==
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By noting the following: <cmath>\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*}</cmath> we see that the answer is <math>\fbox{C}</math>.
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We note that <math>ABFG</math> is a parallelogram because <math>AG = BF = 2</math> and <math>AG \parallel BF</math>. Using the same reasoning, <math>GFCD</math> is also a parallelogram.
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Assume that the height of parallelogram <math>ABFG</math> with respect to base <math>AB</math> is <math>x</math>. Then, the area of parallelogram <math>ABFG</math> is <math>AB * x</math>. The area of triangle <math>EFG</math> is <math>\frac{AB * x}{2}</math>, which is half of the area of parallelogram <math>ABFG</math>.
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Likewise, the area of triangle <math>FGH</math> is half the area of parallelogram <math>GFCD</math>.
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Thus, <math>[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}</math>
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== See also ==
 
== See also ==
 
{{AHSME box|year=1992|num-b=23|num-a=25}}   
 
{{AHSME box|year=1992|num-b=23|num-a=25}}   

Latest revision as of 18:17, 4 September 2022

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

See also

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

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