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 16:34, 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
