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

(See Also)
(5 intermediate revisions by 4 users not shown)
Line 3: Line 3:
 
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 10, and the distance from the center of the circle to the chord is 5. What is the area of the circle?
  
(A) <math>25\pi</math>      (B) <math>50\pi</math>      (C) <math>75\pi</math>      (D) <math>100\pi</math>    (E) <math>125\pi</math>
+
<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==
 
+
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}}
{{MMA Notice}}
+
{{MAA Notice}}
 +
 
 +
[[Category:Introductory Geometry Problems]]

Revision as of 17:23, 18 June 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?

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

Solution

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 \[r^2 = 5^2 + 5^2 = 50\] The area of a circle is $\pi r^2$, so the answer is $\boxed{\text{(B)} 50\pi}$

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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