Difference between revisions of "2017 AMC 12B Problems/Problem 18"
(→Solution 2: Similar triangles) |
|||
Line 54: | Line 54: | ||
Thus the area of <math>\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | Thus the area of <math>\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | |||
+ | Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation: | ||
+ | |||
+ | Draw BF// ED with D on AE. AE=5×(4/7)=20/7. | ||
+ | [ABF]=2×20/7=40/7. | ||
+ | AC:CB:CF=49:35:25. (7/5 ration applied twice) | ||
+ | [ABC]=49/(49+25)[ABF]=140/37. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:43, 19 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 1
Let be the center of the circle. Note that . However, by Power of a Point, , so . Now . Since .
Solution 2: Similar triangles
is the diameter of the circle, so is a right angle, and therefore by AAA similarity, .
Because of this, , so .
Likewise, , so .
Thus the area of .
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:
Draw BF// ED with D on AE. AE=5×(4/7)=20/7. [ABF]=2×20/7=40/7. AC:CB:CF=49:35:25. (7/5 ration applied twice) [ABC]=49/(49+25)[ABF]=140/37.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.