Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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- StarryNight7210 | - StarryNight7210 | ||
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+ | ==Solution 4== | ||
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+ | First, realize that <math>ABCD</math> is not a square. Let <math>O</math> be the midpoint of <math>FE</math>. Since <math>FE=9+9+16=34</math>, we have <math>OF=OE=\frac{34}{2}=17=OB</math> because they are all radii. Since <math>O</math> is also the midpoint of <math>AD</math>, we have <math>OA=\frac{16}2=8</math>. By the Pythagorean Theorem on <math>\triangle BAO</math>, we find that <math>AB=15</math>. The answer is then <math>16\cdot 15=\textbf{(A) }240</math>. | ||
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+ | -franzliszt | ||
==See also== | ==See also== |
Revision as of 15:01, 18 November 2020
Rectangle is inscribed in a semicircle with diameter as shown in the figure. Let and let What is the area of
Solution
First, realize is not a square. It can easily be seen that the diameter of the semicircle is , so the radius is . Express the area of Rectangle as , where . Notice that by the Pythagorean theorem . Then, the area of Rectangle is equal to . ~icematrix
Solution 2
We have , as it is a radius, and since it is half of . This means that . So
~yofro
Solution 3 (coordinate bashing)
Let the midpoint of segment be the origin. Evidently, point is at and is at . Since points and share x-coordinates with and , respectively, we can just find the y-coordinate of (which is just the width of the rectangle) and multiply this by , or . Since the radius of the semicircle is , or , the equation of the circle that our semicircle is a part of is . Since we know that the x-coordinate of is , we can plug this into our equation to obtain that . Since , as the diagram suggests, we know that the y-coordinate of is . Therefore, our answer is , or .
NOTE: The synthetic solution is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier solution.
- StarryNight7210
Solution 4
First, realize that is not a square. Let be the midpoint of . Since , we have because they are all radii. Since is also the midpoint of , we have . By the Pythagorean Theorem on , we find that . The answer is then .
-franzliszt
See also
2020 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
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