Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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<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 | + | ==Solution== |
+ | ===Solution 1=== | ||
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>. | 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>. | ||
Revision as of 20:04, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.