Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 12B Problems/Problem 4"

(→‎Problem)
(→‎Solution)
 
(7 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==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?
  
 
<math>
 
<math>
Line 12: Line 12:
  
 
==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 triangle is <math>\pi r^2</math>, so the answer is <math>\boxed{\text{(B)} 50\pi}</math>
+
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==
 
==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}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 21:06, 27 May 2023

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?

$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$

Solution

Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ 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 $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=\boxed{\textbf{(B) }50\pi}$.

~MRENTHUSIASM

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/bJvXMkwjrtE

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_4&oldid=193602"