2023 AMC 10B Problems/Problem 3: Difference between revisions

Latest revision as of 19:19, 15 November 2023

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-==Problem==
+#redirect[[2023 AMC 12B Problems/Problem 3]]
-A <math>3-4-5</math> right triangle is inscribed in circle <math>A</math>, and a <math>5-12-13</math> right triangle is inscribed in circle <math>B</math>. What is the ratio of the area of circle <math>A</math> to the area of circle <math>B</math>?
-<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==
-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>.
-Therefore the ratio of the areas equals the radius of circle <math>A</math> squared : the radius of circle <math>B</math> squared
-<math>=</math> <math>0.5\times</math> the diameter of circle <math>A</math>, squared : <math>0.5\times</math> the diameter of circle <math>B</math>, squared
-<math>=</math> the diameter of circle <math>A</math>, squared: the diameter of circle <math>B</math>, squared <math>=\boxed{\textbf{(D) }\frac{25}{169}}.</math>
-~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{(D) }\frac{25}{169}}.</math>
-~vsinghminhas