Art of Problem Solving

2012 AMC 12B Problems/Problem 2: Difference between revisions

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== Problem==
== 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?
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>


<math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200</math>
<math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200</math>


 
== Solution ==
 
If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20= \boxed{\textbf{(E)}\ 200}.</math>
==Solution==
If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20=200</math>, <math>\boxed{\text{E}}</math>


== See Also ==
== See Also ==


{{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}
{{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}
{{MAA Notice}}

Latest revision as of 00:07, 19 October 2020

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 $20$. $10\times20= \boxed{\textbf{(E)}\ 200}.$

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
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

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