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

m (Added a little more detail to the solution)
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\text{(E) } 6</math>
 
\text{(E) } 6</math>
  
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== Solution ==
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== Solution 1 ==
 
<math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> 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).
 
<math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> 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: <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>.
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1992|num-b=23|num-a=25}}   
 
{{AHSME box|year=1992|num-b=23|num-a=25}}   

Revision as of 06:29, 28 June 2018

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 (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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