Difference between revisions of "2008 AMC 10A Problems/Problem 17"
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Three rectangles are formed that are <math>3x6</math>. <math>3x3x6=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>3x6</math>. <math>3x3x6=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>. | ||
− | <math>54+9pi \rightarrow</math> | + | <math>54+9pi \rightarrow</math> B |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2008|ab=A|num-b=16|num-a=18}} |
Revision as of 17:08, 15 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: http://i270.photobucket.com/albums/jj87/AndroHM/AMC10A-2008-17.jpg
Three rectangles are formed that are . . degrees = the circle's area. circle's area, .
B
See also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |