Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 8 Problems/Problem 20"

(Created page with "==Problem== ==Solution== ==See Also== {{AMC8 box|year=2013|before=First Problem|num-a=2}} {{MAA Notice}}")
 
(→‎Video Solution 2)
 
(26 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
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>
 +
  
 
==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>.
 +
 +
==Video Solution==
 +
https://youtu.be/tdh0u9_xjN0 ~savannahsolver
  
 
==See Also==
 
==See Also==
−
{{AMC8 box|year=2013|before=First Problem|num-a=2}}
+
{{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

Problem

A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

$\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3$


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, $\sqrt{1^2+1^2}=\sqrt{2}$. The area is $\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}$.

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. AMC logo.png

Thank You for reading these answers by the followers of AoPS.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_20&oldid=222732"