Difference between revisions of "2023 AMC 10B Problems/Problem 3"
Mintylemon66 (talk | contribs) (→Solution) |
Vsinghminhas (talk | contribs) (→Solution) |
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<math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}</math> | <math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}</math> | ||
− | ==Solution== | + | ==Solution 1== |
Since the arc angle of the diameter of a circle is <math>90</math> degrees, the hypotenuse of each these two triangles is respectively the diameter of circles <math>A</math> and <math>B</math>. | Since the arc angle of the diameter of a circle is <math>90</math> degrees, the hypotenuse of each these two triangles is respectively the diameter of circles <math>A</math> and <math>B</math>. | ||
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~Mintylemon66 | ~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 <math>\frac{5^2}{13^2} =\boxed{\textbf{(B) }\frac{25}{169}}.</math> | ||
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+ | ~vsinghminhas |
Revision as of 16:12, 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