Difference between revisions of "2008 AMC 10A Problems/Problem 17"

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Three rectangles are formed that are <math>3*6</math>.  <math>3*3*6=54</math>.  <math>360-90-90-60=120</math> degrees = <math>1/3</math> the circle's area.  <math>1/3*3= 1</math> circle's area, <math>9 \pi</math>.   
 
Three rectangles are formed that are <math>3*6</math>.  <math>3*3*6=54</math>.  <math>360-90-90-60=120</math> degrees = <math>1/3</math> the circle's area.  <math>1/3*3= 1</math> circle's area, <math>9 \pi</math>.   

Revision as of 07:12, 16 June 2008

Problem

An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point of the triangle?

$\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\mathrm{(E)}\ 9\left(\sqrt{3}{+1\right)^2\pi$ (Error compiling LaTeX. Unknown error_msg)

Solution

Diagram: AMC10A-2008-17.png

Three rectangles are formed that are $3*6$. $3*3*6=54$. $360-90-90-60=120$ degrees = $1/3$ the circle's area. $1/3*3= 1$ circle's area, $9 \pi$.

$54+9pi \rightarrow$ B

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions