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Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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==Problem==
 
The diameter <math>\overline{AB}</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>\overline{AE}</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?
 
The diameter <math>\overline{AB}</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>\overline{AE}</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?
  
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Answer is D
 
Answer is D
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==See Also==
 
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{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 
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{{MAA Notice}}
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:53, 16 February 2017

Problem

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$

Solution

Answer is D

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 10 Problems and Solutions

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