Difference between revisions of "2023 AMC 10B Problems/Problem 3"

(Solution)
Line 11: Line 11:
 
Therefore the ratio of the areas equals the radius of circle <math>A</math> squared : the radius of circle <math>B</math> squared
 
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> <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>=</math> the diameter of circle <math>A</math>, squared: the diameter of circle <math>B</math>, squared <math>=\boxed{\textbf{(B) }\frac{25}{169}}.</math>  
<math>=\boxed{\textbf{(B) }\dfrac{25}{169}.</math>  
 
 
 
 
~Mintylemon66
 
~Mintylemon66

Revision as of 14:55, 15 November 2023

Problem

A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. What is the ratio of the area of circle $A$ to the area of circle $B$?


$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}$

Solution

Since the arc angle of the diameter of a circle is $90$ degrees, the hypotenuse of each these two triangles is respectively the diameter of circles $A$ and $B$.

Therefore the ratio of the areas equals the radius of circle $A$ squared : the radius of circle $B$ squared $=$ $0.5\times$ the diameter of circle $A$, squared : $0.5\times$ the diameter of circle $B$, squared $=$ the diameter of circle $A$, squared: the diameter of circle $B$, squared $=\boxed{\textbf{(B) }\frac{25}{169}}.$ ~Mintylemon66