2013 AMC 8 Problems/Problem 20: Difference between revisions
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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== | ||
<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>. | 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>. | ||
==List of steps== | |||
1. Draw a picture, realize the radius comes from the midpoint of rectangle | |||
2. Using the pythagorean theorm, solve for the radius with is square root of 2 | |||
3. Solve for the area of the circle using the radius pi times radius squared, with will give you 2pi | |||
4. Divide by 2, getting pi | |||
-Jason Da Goat | |||
==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 13:52, 24 August 2025
Problem
A 
rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
Solution
![[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]](http://latex.artofproblemsolving.com/d/c/6/dc675cf6a5605847c1c23c56fbe9b8a37ad43fd2.png)
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 
.
List of steps
1. Draw a picture, realize the radius comes from the midpoint of rectangle 2. Using the pythagorean theorm, solve for the radius with is square root of 2 3. Solve for the area of the circle using the radius pi times radius squared, with will give you 2pi 4. Divide by 2, getting pi -Jason Da Goat
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
Thank You for reading these answers by the followers of AoPS.
