Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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[EIDJ]&=[EIJ]+[JDI] \ | [EIDJ]&=[EIJ]+[JDI] \ | ||
&=\frac{1}{2}(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}) + \frac{1}{2} (\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \ | &=\frac{1}{2}(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}) + \frac{1}{2} (\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \ | ||
− | &=\frac{1}{2}(\frac{1}{3})+\frac{1}{2}( | + | &=\frac{1}{2}(\frac{1}{3})+\frac{1}{2}(\frac{1}{4}-\frac{1}{12}) \ |
&=\frac{1}{6}+\frac{1}{12} \ | &=\frac{1}{6}+\frac{1}{12} \ | ||
&=\frac{1}{4} \ | &=\frac{1}{4} \ |
Revision as of 16:14, 15 July 2024
Problem 14
The vertex of a square is at the center of square The length of a side of is and the length of a side of is Side intersects at and intersects at If angle the area of quadrilateral is
Solution 1
Draw a diagram. Split quadrilateral into and Let the perpendicular from point intersect at , and let the perpendicular from point intersect at We know because since is a square, as given, and so Since is at the center of square , By Additionally, we know so and we know so From here, we can sum the areas of and to get the area of quadrilateral Therefore,
Thus, our answer is
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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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