Difference between revisions of "2023 AMC 12B Problems/Problem 3"
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<math>\textbf{(A) }\frac{9}{25}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{1}{5}\qquad\textbf{(D) }\frac{25}{169}\qquad\textbf{(E) }\frac{4}{25}</math> | <math>\textbf{(A) }\frac{9}{25}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{1}{5}\qquad\textbf{(D) }\frac{25}{169}\qquad\textbf{(E) }\frac{4}{25}</math> | ||
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| + | ==Solution 1== | ||
| + | Because the triangle are right triangles, we know the hypotenuses are diameters of circles <math>A</math> and <math>B</math>. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply <math>\pi</math> to get <math>6.25\pi</math> and <math>42.25\pi</math> as the areas of the circles. Multiply 4 on both numbers to get <math>25\pi</math> and <math>169\pi</math>. Lastly, divide, to get your answer: <math>\frac{25}{169}</math> = <math>\boxed{\textbf{(D) \frac{25}{169}}</math>. | ||
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| + | ~zhenghua | ||
| + | ~Failure.net | ||
Revision as of 13:25, 15 November 2023
Problem
A 3-4-5 right triangle is inscribed circle 
, and a 5-12-13 right triangle is inscribed in circle 
. What is the ratio of the area of circle to circle ?
Solution 1
Because the triangle are right triangles, we know the hypotenuses are diameters of circles and . Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply 
to get 
and 
as the areas of the circles. Multiply 4 on both numbers to get 
and 
. Lastly, divide, to get your answer: 
= $\boxed{\textbf{(D) \frac{25}{169}}$ (Error compiling LaTeX. Unknown error_msg).
~zhenghua ~Failure.net
