Difference between revisions of "2018 AMC 12B Problems/Problem 4"
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==Problem== | ==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? | + | A circle has a chord of length <math>10</math>, and the distance from the center of the circle to the chord is <math>5</math>. What is the area of the circle? |
| − | (A) | + | <math> |
| + | \textbf{(A) }25\pi \qquad | ||
| + | \textbf{(B) }50\pi \qquad | ||
| + | \textbf{(C) }75\pi \qquad | ||
| + | \textbf{(D) }100\pi \qquad | ||
| + | \textbf{(E) }125\pi \qquad | ||
| + | </math> | ||
==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. | ||
| + | <asy> | ||
| + | /* Made by MRENTHUSIASM */ | ||
| + | size(200); | ||
| + | pair O, A, B, M; | ||
| + | O = (0,0); | ||
| + | A = (-5,5); | ||
| + | B = (5,5); | ||
| + | M = midpoint(A--B); | ||
| + | draw(Circle(O,5sqrt(2))); | ||
| + | dot("$O$", O, 1.5*S, linewidth(4.5)); | ||
| + | dot("$A$", A, 1.5*NW, linewidth(4.5)); | ||
| + | dot("$B$", B, 1.5*NE, linewidth(4.5)); | ||
| + | dot("$M$", M, 1.5*N, linewidth(4.5)); | ||
| + | draw(A--B^^M--O^^A--O^^M--O^^B--O); | ||
| + | label("$5$", midpoint(A--M), 1.5*N); | ||
| + | label("$5$", midpoint(B--M), 1.5*N); | ||
| + | label("$5$", midpoint(O--M), 1.5*E); | ||
| + | label("$r$", midpoint(O--A), 1.5*SW); | ||
| + | label("$r$", midpoint(O--B), 1.5*SE); | ||
| + | </asy> | ||
| + | Note 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>. | ||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
| + | https://youtu.be/bJvXMkwjrtE | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==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}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 21:06, 27 May 2023
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.
![[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, M; O = (0,0); A = (-5,5); B = (5,5); M = midpoint(A--B); draw(Circle(O,5sqrt(2))); dot("$O$", O, 1.5*S, linewidth(4.5)); dot("$A$", A, 1.5*NW, linewidth(4.5)); dot("$B$", B, 1.5*NE, linewidth(4.5)); dot("$M$", M, 1.5*N, linewidth(4.5)); draw(A--B^^M--O^^A--O^^M--O^^B--O); label("$5$", midpoint(A--M), 1.5*N); label("$5$", midpoint(B--M), 1.5*N); label("$5$", midpoint(O--M), 1.5*E); label("$r$", midpoint(O--A), 1.5*SW); label("$r$", midpoint(O--B), 1.5*SE); [/asy]](http://latex.artofproblemsolving.com/a/9/8/a98a012a46954b990fd2592abb76aa48f7e6cef5.png)
Note that 
Since 
we conclude that 
and 
are congruent isosceles right triangles. It follows that 
so the area of 
is 
.
~MRENTHUSIASM
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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. 
