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

Revision as of 20:29, 17 February 2013

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 $\boxed{\textbf{(E)}\ 200}$

@@ Line 13: / Line 13: @@
 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>.
-Therefore the area is <math>20\cdot 10 = 200</math> or \boxed{\textbf{(E)}\ 200}$
+Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{\textbf{(E)}\ 200}</math>

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

Revision as of 20:29, 17 February 2013

Problem

Solution