2023 AMC 12B Problems/Problem 3

Revision as of 13:29, 15 November 2023 by Failure.net (talk | contribs) (Solution 1)

Problem

A 3-4-5 right triangle is inscribed 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 circle $B$?

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


Solution 1

Because the triangle are right triangles, we know the hypotenuses are diameters of circles $A$ and $B$. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply $\pi$ to get $6.25\pi$ and $42.25\pi$ as the areas of the circles. Multiply 4 on both numbers to get $25\pi$ and $169\pi$. Cancel out the $\pi$, and lastly, divide, to get your answer: $\boxed{\frac{25}{169}}$.


~Failure.net