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

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==Problem==
 
==Problem==
  
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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?
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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>
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==Solution==
 
==Solution==
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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>
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The shortest lines segment from 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> so the area of the circle is <math>\pi r^2=\boxed{\textbf{(B) }50\pi}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:45, 18 September 2021

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 lines 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 $\pi r^2=\boxed{\textbf{(B) }50\pi}$.

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

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