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

m (→‎Solution 2)
m (→‎Solution 2)
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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>
 
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>\sqrt{2}^2</math> * <math>\pi</math> = 2<math>\pi</math>
+
The area of the circle is <math>\left ( \sqrt{2} \right ) ^2 \cdot \pi</math> = 2<math>\pi</math>
  
 
Finally, the area of the semicircle is <math>\pi</math>, so the answer is <math>\boxed{C}</math>.
 
Finally, the area of the semicircle is <math>\pi</math>, so the answer is <math>\boxed{C}</math>.

Revision as of 14:11, 6 October 2020

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 $\left ( \sqrt{2} \right ) ^2 \cdot \pi$ = 2$\pi$

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

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

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