Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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==Solution== | ==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 17:30, 17 March 2008
Problem
Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?
Solution
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles and .
We know that triangles and are similar because . Also, since , the ratio of the distance from to to the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from to is .
The area of is simply , the area of is , and the area of rectangle is .
Taking the area of rectangle and subtracting the combined area of and yields .
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 |