2012 AMC 12B Problems/Problem 2

Problem

A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle? $[asy]draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); draw(circle((10,5),5));[/asy]$ $\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200$

Solution

If the radius is $5$ , then the width is $10$ , hence the length is $20$ . $10\times20= \boxed{\textbf{(E)}\ 200}.$

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2012 AMC 12B Problems/Problem 2

Problem

Solution

See Also