Difference between revisions of "2013 AMC 8 Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | A <math>1\times 2</math> rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle? | + | A <math>1\times 2</math> rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle? |
<math>\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3</math> | <math>\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3</math> | ||
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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, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}</math>. | 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, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}</math>. | ||
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+ | ==Solution 2== | ||
+ | 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> | ||
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+ | The area of the circle is <math>\left ( \sqrt{2} \right ) ^2 \pi = 2\pi</math> | ||
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+ | Finally, the area of the semicircle is <math>\pi</math>, so the answer is <math>\boxed{C}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=19|num-a=21}} | {{AMC8 box|year=2013|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | Thank You for reading these answers by the followers of AoPS. |
Latest revision as of 15:13, 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
Finally, the area of the semicircle is , so the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Thank You for reading these answers by the followers of AoPS.