2013 AMC 8 Problems/Problem 20

Revision as of 19:31, 27 November 2013 by Fadebekun (talk | contribs) (Solution)

Problem

A $1\times 2$ rectangle is inscribed in a semicircle with 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 $\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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