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Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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==Solution==
 
==Solution==
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{{solution}}
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We can obtain the solution by calculating the area of rectangle <math>ABGH</math> minus the combined area of triangles <math>\triangle AHG</math> and <math>\triangle CGM</math>.
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We know that triangles <math>\triangle AMH</math> and <math>\triangle CGM</math> are similar because <math>\overline{AH} \parallel \overline{CG}</math>. Also, since <math>\frac{AH}{CG} = \frac{3}{2}</math>, the ratio of the distance from <math>M</math> to <math>\overline{AH}</math> to the distance from <math>M</math> to <math>\overline{CG}</math> is also <math>\frac{3}{2}</math>. Solving with the fact that the distance from <math>\overline{AH}</math> to <math>\overline{CG}</math> is 4, we see that the distance from <math>M</math> to <math>\overline{CG}</math> is <math>\frac{8}{5}</math>.
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The area of <math>\triangle AHG</math> is simply <math>\frac{1}{2} \cdot 4 \cdot 12 = 24</math>, the area of <math>\triangle CGM</math> is <math>\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}</math>, and the area of rectangle <math>ABGH</math> is <math>4 \cdot 12 = 48</math>.
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Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}</math>.
  
 
==See also==
 
==See also==

Revision as of 18:30, 17 March 2008

Problem

Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?

2007-AMC-10A--18.png

$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$

Solution

We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $\triangle AHG$ and $\triangle CGM$.

We know that triangles $\triangle AMH$ and $\triangle CGM$ are similar because $\overline{AH} \parallel \overline{CG}$. Also, since $\frac{AH}{CG} = \frac{3}{2}$, the ratio of the distance from $M$ to $\overline{AH}$ to the distance from $M$ to $\overline{CG}$ is also $\frac{3}{2}$. Solving with the fact that the distance from $\overline{AH}$ to $\overline{CG}$ is 4, we see that the distance from $M$ to $\overline{CG}$ is $\frac{8}{5}$.

The area of $\triangle AHG$ is simply $\frac{1}{2} \cdot 4 \cdot 12 = 24$, the area of $\triangle CGM$ is $\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}$, and the area of rectangle $ABGH$ is $4 \cdot 12 = 48$.

Taking the area of rectangle $ABGH$ and subtracting the combined area of $\triangle AHG$ and $\triangle CGM$ yields $48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}$.

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
All AMC 10 Problems and Solutions
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