Difference between revisions of "2023 AMC 10B Problems/Problem 3"
Mintylemon66 (talk | contribs) (→Solution) |
Mintylemon66 (talk | contribs) (→Solution) |
||
| Line 12: | Line 12: | ||
<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>=\boxed{\textbf{(B) }\frac{25}{169}}.</math> | <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> | ||
| + | |||
| + | |||
~Mintylemon66 | ~Mintylemon66 | ||
Revision as of 14:55, 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
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
