Difference between revisions of "2018 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | The shortest | + | The shortest line segment from 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> so the area of the circle is <math>\pi r^2=\boxed{\textbf{(B) }50\pi}</math>. |
==See Also== | ==See Also== | ||
Revision as of 10:58, 19 September 2021
Problem
A circle has a chord of length 
, and the distance from the center of the circle to the chord is 
. What is the area of the circle?
Solution
The shortest line segment from 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/7/2/f/72f1f7169ae54f4d9cb3c76f6aef4e1f2627153c.png)
so the area of the circle 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. 
