Difference between revisions of "2018 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | + | Let <math>O</math> be the center of the circle, <math>\overline{AB}</math> be the chord, and <math>M</math> be the midpoint of <math>\overline{AB},</math> as shown below. | |
| + | |||
| + | <b>DIAGRAM NEEDED</b> | ||
| + | |||
| + | Recall that <math>\overline{OM}\perp\overline{AB}.</math> Since <math>OM=AM=BM=5,</math> we conclude that <math>\triangle OMA</math> and <math>\triangle OMB</math> are congruent isosceles right triangles. It follows that <math>r=5\sqrt2,</math> so the area of <math>\odot O</math> is <math>\pi r^2=\boxed{\textbf{(B) }50\pi}</math>. | ||
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| + | ~MRENTHUSIASM | ||
==See Also== | ==See Also== | ||
Revision as of 08:30, 20 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
Let 
be the center of the circle, 
be the chord, and 
be the midpoint of 
as shown below.
DIAGRAM NEEDED
Recall that 
Since 
we conclude that 
and 
are congruent isosceles right triangles. It follows that 
so the area of 
is 
.
~MRENTHUSIASM
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. 
