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?
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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>.
 +
 
 +
==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>
 +
 
 +
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>.
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=6WPBluEpmMA
 +
 
 +
https://youtu.be/tdh0u9_xjN0 ~savannahsolver
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/0g14IJJ2Z-8 Soo, DRMS, NM
 +
 
  
 
==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.

Revision as of 18:01, 5 May 2022

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 $\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}$.

Solution 2

Double the figure to get a square with side length $2$. The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is $\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$. The circle’s radius ,therefore, is $\sqrt{2}$

The area of the circle is $\left ( \sqrt{2} \right ) ^2 \pi = 2\pi$

Finally, the area of the semicircle is $\pi$, so the answer is $\boxed{C}$.


Video Solution

https://www.youtube.com/watch?v=6WPBluEpmMA

https://youtu.be/tdh0u9_xjN0 ~savannahsolver

Video Solution 2

https://youtu.be/0g14IJJ2Z-8 Soo, DRMS, NM


See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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.