Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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==Solution 6== | ==Solution 6== | ||
A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort. | A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort. | ||
− | + | ==Solution 7== The other side will be <math>\frac{16+9}{9}=\frac{25} which we know it is a </math>AB=15<math>. or so </math>16\cdot 15=\textbf{(A) }240$. | |
− | == | + | ~oceanxia |
{{AMC8 box|year=2020|num-b=17|num-a=19}} | {{AMC8 box|year=2020|num-b=17|num-a=19}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:16, 19 November 2020
Rectangle is inscribed in a semicircle with diameter as shown in the figure. Let and let What is the area of
Contents
Solution 1
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
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: .
Solution 6
A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort. ==Solution 7== The other side will be AB=1516\cdot 15=\textbf{(A) }240$. ~oceanxia
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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