Difference between revisions of "2013 AMC 8 Problems/Problem 20"
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Double the figure to get a square with side length <math>2</math>. The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is <math>\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}</math>. The circle’s radius ,therefore, is <math>\sqrt{2}</math> | Double the figure to get a square with side length <math>2</math>. The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is <math>\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}</math>. The circle’s radius ,therefore, is <math>\sqrt{2}</math> | ||
− | The area of the circle is <math>\left ( \sqrt{2} \right ) ^2 \pi</math> = 2 | + | The area of the circle is <math>\left ( \sqrt{2} \right ) ^2 \pi</math> = 2\pi<math> |
− | Finally, the area of the semicircle is <math>\pi< | + | Finally, the area of the semicircle is </math>\pi<math>, so the answer is </math>\boxed{C}$. |
==See Also== | ==See Also== |
Revision as of 14:12, 6 October 2020
Contents
Problem
A rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
Solution
A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, . The area is .
Solution 2
Double the figure to get a square with side length . The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is . The circle’s radius ,therefore, is
The area of the circle is = 2\pi\pi\boxed{C}$.
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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