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

(List of steps)
(List of steps)
 
Line 46: Line 46:
 
3. Solve for the area of the circle using the radius pi times radius squared, with will give you 2pi
 
3. Solve for the area of the circle using the radius pi times radius squared, with will give you 2pi
 
4. Divide by 2, getting pi
 
4. Divide by 2, getting pi
 +
-Jason Da Goat
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:52, 24 August 2025

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

List of steps

1. Draw a picture, realize the radius comes from the midpoint of rectangle 2. Using the pythagorean theorm, solve for the radius with is square root of 2 3. Solve for the area of the circle using the radius pi times radius squared, with will give you 2pi 4. Divide by 2, getting pi -Jason Da Goat

Video Solution

https://youtu.be/tdh0u9_xjN0 ~savannahsolver

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

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.