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> | ||
+ | |||
==Solution== | ==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, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{(C) \pi}</math>. | + | <asy> |
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; usepackage("amsmath"); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = 2.392515856236789, xmax = 4.844947145877386, ymin = 6.070697674418619, ymax = 8.062241014799170; /* image dimensions */ | ||
+ | pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); | ||
+ | |||
+ | draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)--cycle, zzttqq); | ||
+ | /* draw figures */ | ||
+ | draw((2.912600422832983,6.903678646934476)--(4.326813985206080,6.903678646934476)); | ||
+ | draw(shift((3.619707204019532,6.903678646934476))*xscale(0.7071067811865487)*yscale(0.7071067811865487)*arc((0,0),1,0.000000000000000,180.0000000000000)); | ||
+ | draw((3.619707204019532,6.903678646934476)--(4.119707204019532,6.903678646934476)); | ||
+ | draw((3.619707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)); | ||
+ | draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476), zzttqq); | ||
+ | draw((4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476), zzttqq); | ||
+ | draw((4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476), zzttqq); | ||
+ | draw((3.119707204019531,6.903678646934476)--(3.119707204019531,7.403678646934482), zzttqq); | ||
+ | label("$1$",(3.847061310782247,6.924820295983102),SE*labelscalefactor); | ||
+ | label("$1$",(4.155729386892184,7.208118393234687),SE*labelscalefactor); | ||
+ | draw((3.619707204019532,6.903678646934476)--(4.119707204019532,7.403678646934476)); | ||
+ | label("$\sqrt{2}$",(3.711754756871041,7.288456659619466),SE*labelscalefactor); | ||
+ | label("$2$",(3.563763213530660,7.563298097251601),SE*labelscalefactor); | ||
+ | /* dots and labels */ | ||
+ | dot((2.912600422832983,6.903678646934476)); | ||
+ | dot((4.326813985206080,6.903678646934476)); | ||
+ | dot((3.619707204019532,6.903678646934476)); | ||
+ | dot((4.119707204019532,6.903678646934476),blue); | ||
+ | dot((3.619707204019532,6.903678646934476)); | ||
+ | dot((3.119707204019531,6.903678646934476),blue); | ||
+ | dot((3.119707204019531,7.403678646934482),blue); | ||
+ | dot((4.119707204019532,7.403678646934476),blue); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/tdh0u9_xjN0 ~savannahsolver | ||
==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 08:51, 16 July 2024
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 .
Video Solution
https://youtu.be/tdh0u9_xjN0 ~savannahsolver
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.