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>\triangle | + | 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>? |
− | 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> | ||
− | + | ==Solution== | |
+ | Answer is D | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Revision as of 09:45, 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 ?
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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