Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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− | We can use coordinates to solve this. Let <math>H=(0,0).</math> Thus, we have <math>A=(0,12), C=(4,8), G=(4,0).</math> Therefore, <math>AG</math> has equation <math>-3x+12=y</math> and <math>HC</math> has equation <math>2x=y.</math> Solving, we have <math>M=(12/5,24/5).</math> Using Shoelace Theorem (or you could connect <math>LC</math> and solve for the resulting triangle + trapezoid areas), we find <math>[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.</math> | + | We can use coordinates to solve this. Let <math>H=(0,0).</math> Thus, we have <math>A=(0,12), C=(4,8), G=(4,0).</math> Therefore, <math>AG</math> has equation <math>-3x+12=y</math> and <math>HC</math> has equation <math>2x=y.</math> Solving, we have <math>M=(12/5,24/5).</math> Using [[Shoelace Theorem]] (or you could connect <math>LC</math> and solve for the resulting triangle + trapezoid areas), we find <math>[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.</math> |
==See also== | ==See also== |
Revision as of 18:56, 25 June 2018
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
Solution 1
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 .
Solution 2
Extend and and call their intersection .
The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is .
The triangles and are similar as well, and we now know that the ratio of their dimensions is .
Draw altitudes from onto and , let their feet be and . We get that . Hence . (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem(http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, must be , by the harmonic mean. Thus, must be .)
Then the area of is , and the area of can be obtained by subtracting the area of , which is . Hence the answer is .
Solution 3
We can use coordinates to solve this. Let Thus, we have Therefore, has equation and has equation Solving, we have Using Shoelace Theorem (or you could connect and solve for the resulting triangle + trapezoid areas), we find
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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