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 line segment from the center of the circle to a chord is the perpendicular bisector of the chord. Applying the Pythagorean Theorem, we find that $r^2 = 5^2 + 5^2 = 50,$ so the area of the circle is $\pi r^2=\boxed{\textbf{(B) }50\pi}$ .

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 4

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