Difference between revisions of "2023 AMC 10B Problems/Problem 3"
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− | <math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{ | + | <math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{15}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{25}{169}\qquad\textbf{(E) }\frac{9}{25}</math> |
==Solution 1== | ==Solution 1== |
Revision as of 17:34, 15 November 2023
Problem
A right triangle is inscribed in circle
, and a
right triangle is inscribed in circle
. What is the ratio of the area of circle
to the area of circle
?
Solution 1
Since the arc angle of the diameter of a circle is degrees, the hypotenuse of each these two triangles is respectively the diameter of circles
and
.
Therefore the ratio of the areas equals the radius of circle squared : the radius of circle
squared
the diameter of circle
, squared :
the diameter of circle
, squared
the diameter of circle
, squared: the diameter of circle
, squared
~Mintylemon66
Solution 2
The ratio of areas of circles is the same as the ratios of the diameters squared. Since this is a right triangle the hypotenuse of each triangle will be the diameter of the circle. This yields the expression
~vsinghminhas