Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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− | The diameter <math>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>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>ABC</math>? | + | The diameter <math>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>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>? | ||
<math>\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}</math> | <math>\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}</math> |
Revision as of 09:22, 16 February 2017
The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?