Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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Sweetmango77 (talk | contribs) (→Solution 4) |
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+ | ==Solution 5 -SweetMango77== | ||
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+ | This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is <math>1+n</math> with the <math>1</math> part at one side, and the <math>n</math> part at the other side, then the height from the end of the <math>1</math> side and the start of the <math>n</math> side is <math>\sqrt{n}</math>. | ||
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+ | Using this, we can scale the image down by <math>9</math> to get what we note: The other side will be <math>\frac{16+9}{9}=\frac{25}{3}=\left(\frac{5}{3}\right)^2</math>. Then, the height of that part will be <math>\frac{5}{3}</math>. But, we have to scale it back up by <math>9</math> to get a height of <math>15</math>. Multiplying by <math>16</math> gives our desired answer: <math>\textbf{(A) }240</math>. | ||
==See also== | ==See also== |
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
Contents
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
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:
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.