2018 AMC 12B Problems/Problem 4

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

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