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

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

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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