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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}}")
 
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==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>.
 +
 +
==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>
 +
 +
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|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.

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 (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. AMC logo.png

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

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