Difference between revisions of "2020 AMC 10A Problems/Problem 20"
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− | Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, and using the fact that <math>cos\angle{AEB}=\frac{1}{\sqrt5}</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>, <math>[ABCD]=[ACD]+\frac{1}{5}[ACD]=300+60=\boxed {\textbf{D) }360}</math> | + | Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, and using the fact that <math>cos\angle{AEB}=\frac{1}{\sqrt5}</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>, <math>[ABCD]=[ACD]+\frac{1}{5}[ACD]=300+60=\boxed {\textbf{(D) }360}</math> |
(This solution is incomplete, can someone complete it please-Lingjun) | (This solution is incomplete, can someone complete it please-Lingjun) |
Revision as of 19:40, 1 February 2020
Contents
Problem
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find to get the area of the whole quadrilateral. Drop an altitude from to and call the point of intersection . Let . Since , then . By dropping this altitude, we can also see two similar triangles, and . Since is , and , we get that . Now, if we redraw another diagram just of , we get that . Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
(I'm very sorry if you're a visual learner)
~Ultraman
Solution 2 (Pro Guessing Strats)
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to and , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , is indeed a square. The area of the triangle is thus 60, so the answer is .
~tigershark22 ~(edited by HappyHuman)
Solution 3 (coordinates)
Let the points be , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discard the solution as should be negative. Thus, we conclude that .
Solution 4 (Trigonometry)
Using the law of cosines, express and in terms of . The sum of these two equations is by the Pythagorean Theorem. Solving for , and using the fact that , we find . Since and , , which is five times ,
(This solution is incomplete, can someone complete it please-Lingjun) Latex edited by kc5170
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2020 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.