Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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<math>\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi</math> | <math>\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi</math> | ||
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| + | ==Solution 1== | ||
| + | Let the center of the first circle be <math>O.</math> By Pythagorean Theorem, | ||
| + | <cmath>AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}</cmath> | ||
| + | Now, notice that since <math>\angle ABO</math> is <math>90</math> degrees, so arc <math>AO</math> is <math>180</math> degrees and <math>AO</math> is the diameter. Thus, the radius is <math>\sqrt{26},</math> so the area is <math>\boxed{26\pi}.</math> | ||
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| + | - kante314 | ||
Revision as of 21:05, 22 November 2021
Isosceles triangle 
has 
, and a circle with radius 
is tangent to line 
at 
and to line 
at 
. What is the area of the circle that passes through vertices 
, , and 
Solution 1
Let the center of the first circle be 
By Pythagorean Theorem,
![\[AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}\]](http://latex.artofproblemsolving.com/f/a/a/faab0344e40b6da42c07a30d261447e99f9aa063.png)
Now, notice that since 
is 
degrees, so arc 
is 
degrees and is the diameter. Thus, the radius is 
so the area is 
- kante314
