Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
− | We note that <math>\triangle ACB ~ \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = (\frac{4}{\sqrt{74}})^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>. | + | We note that <math>\triangle ACB ~ \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \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 13:11, 16 February 2017
Contents
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
Notice that and
are right triangles. Then
.
, so
. We also find that
, and thus the area of
is
.
Solution 2
We note that by
similarity. Also, since the area of
and
,
, so the area of
.
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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