Difference between revisions of "2018 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | The shortest segment that connects the center of the circle to a chord is the perpendicular bisector of the chord. Applying the Pythagorean theorem, we find that <cmath>r^2 = 5^2 + 5^2 = 50</cmath> The area of a | + | The shortest segment that connects the center of the circle to a chord is the perpendicular bisector of the chord. Applying the Pythagorean theorem, we find that <cmath>r^2 = 5^2 + 5^2 = 50</cmath> The area of a circle is <math>\pi r^2</math>, so the answer is <math>\boxed{\text{(B)} 50\pi}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2018|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:59, 1 March 2018
Problem
A circle has a chord of length 10, and the distance from the center of the circle to the chord is 5. What is the area of the circle?
Solution
The shortest segment that connects the center of the circle to a chord is the perpendicular bisector of the chord. Applying the Pythagorean theorem, we find that ![\[r^2 = 5^2 + 5^2 = 50\]](http://latex.artofproblemsolving.com/d/e/f/def9253c749de80deb042412eb5965c8558b8c5c.png)
The area of a circle is 
, so the answer is 
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
