Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | First, realize <math>ABCD</math> is not a square. It can easily be seen that the diameter of the semicircle is <math>9+16+9=34</math>, so the radius is <math>\frac{34}{2}=17</math>. Express the area of Rectangle <math>ABCD</math> as <math>16h</math>, where <math>h=AB</math>. Notice that by the Pythagorean theorem <math>8^2+h^{2}=17^{2}\implies h=15</math>. Then, the area of Rectangle <math>ABCD</math> is equal to <math>\boxed{\textbf{(A) }240}</math>. ~icematrix | + | First, realize <math>ABCD</math> is not a square. It can easily be seen that the diameter of the semicircle is <math>9+16+9=34</math>, so the radius is <math>\frac{34}{2}=17</math>. Express the area of Rectangle <math>ABCD</math> as <math>16h</math>, where <math>h=AB</math>. Notice that by the Pythagorean theorem <math>8^2+h^{2}=17^{2}\implies h=15</math>. Then, the area of Rectangle <math>ABCD</math> is equal to <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>. ~icematrix |
+ | ==See also== | ||
{{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 00:57, 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
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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