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

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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 triangle is <math>\pi r^2</math>, so the answer is <math>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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:43, 16 February 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?

(A) $25\pi$ (B) $50\pi$ (C) $75\pi$ (D) $100\pi$ (E) $125\pi$

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 triangle is $\pi r^2$, so the answer is $B) 50\pi$

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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