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? | + | 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> | ||
Revision as of 13:28, 14 April 2019
Contents
Problem
A 
rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
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]](http://latex.artofproblemsolving.com/d/c/6/dc675cf6a5605847c1c23c56fbe9b8a37ad43fd2.png)
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, 
. The area is 
.
Solution 2
Double the figure to get a square with side length 
. The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is 
=
=2
. The circle’s radius ,therefore, is
The area of the circle is 
* 
= 2
Finally, the area of the semicircle is , so the answer is 
.
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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