Difference between revisions of "2012 AMC 10B Problems/Problem 2"

(Problem)
m (Problem)
 
(5 intermediate revisions by 3 users not shown)
Line 8: Line 8:
  
 
<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>
 +
[[Category: Introductory Geometry Problems]]
  
 
==Solution==
 
==Solution==
Line 13: Line 14:
 
Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</math>.
 
Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</math>.
 
Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>.
 
Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>.
Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{E}</math>.
+
Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{\textbf{(E)}\ 200}</math>
 +
 
 +
==See Also==
 +
 
 +
{{AMC10 box|year=2012|ab=B|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Latest revision as of 11:46, 13 August 2014

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

Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$, the diameter is $2\cdot 5 = 10$. Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\cdot 10 = 20$. Therefore the area is $20\cdot 10 = 200$ or $\boxed{\textbf{(E)}\ 200}$

See Also

2012 AMC 10B (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 10 Problems and Solutions

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