Difference between revisions of "2023 AMC 10B Problems/Problem 3"
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| + | ==Problem== | ||
| + | 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{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | 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{(B) }\dfrac{25}{169}.</math> | ||
| + | |||
| + | ~Mintylemon66 | ||
Revision as of 14:52, 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
$=\boxed{\textbf{(B) }\dfrac{25}{169}.$ (Error compiling LaTeX. Unknown error_msg)
~Mintylemon66