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

(Created page with "== Problem == Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segmen...")
 
(Added a solution)
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== Problem ==
 
== Problem ==
  
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Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segments <math>\overline{AB},\overline{BC}</math> and <math>\overline{AD}</math>, respectively, with <math>AE=BF=AG=2</math>. Let the line through <math>G</math> parallel to <math>\overline{EF}</math> intersect <math>\overline{CD}</math> at <math>H</math>. The area of quadrilateral <math>EFGH</math> is
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Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segments <math>\overline{AB},\overline{BC}</math> and <math>\overline{AD}</math>, respectively, with <math>AE=BF=AG=2</math>. Let the line through <math>G</math> parallel to <math>\overline{EF}</math> intersect <math>\overline{CD}</math> at <math>H</math>. The area of quadrilateral <math>EFHG</math> is
  
 
<math>\text{(A) } 4\quad
 
<math>\text{(A) } 4\quad
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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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<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>\|pxq\|=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.
  
 
== See also ==
 
== See also ==

Revision as of 02:20, 20 February 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

$\fbox{C}$ Use vectors. Place an origin at $A$, with $B = p, D = q, C = p + q$. We know that $\|pxq\|=10$, 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.

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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