Difference between revisions of "2020 AMC 8 Problems/Problem 18"
(→Solution 3 (coordinate bashing))
|Line 31:||Line 31:|
Revision as of 17:59, 18 November 2020
Rectangle is inscribed in a semicircle with diameter as shown in the figure. Let and let What is the area of
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
We have , as it is a radius, and since it is half of . This means that . So
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.
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 .
Solution 5 -SweetMango77
This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is with the part at one side, and the part at the other side, then the height from the end of the side and the start of the side is .
Using this, we can scale the image down by to get what we note: The other side will be . Then, the height of that part will be . But, we have to scale it back up by to get a height of . Multiplying by gives our desired answer: .
|2020 AMC 8 (Problems • Answer Key • Resources)|
|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|