Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | + | Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math>, and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>. | |
==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 12:57, 16 February 2017
Problem
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
Notice that 
and 
are right triangles. Then 
. 
, so 
. We also find that 
, and thus the area of is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
