The formula for the surface of a sphere is \(S=4πr^2\). Two lead balls have radii 3 inches and 6 inches, respectively. A third ball has surface area equal to the sum of the surface areas of the two smaller balls. What is the radius, in inches, of the third ball?

(A) \(3\sqrt{10}\)
(B) \(3\sqrt{5}\)
(C) \(36π\)
(D) \(45π\)
(E) \(180π\)

It's not really a hard question, but I got it wrong because I didn't take enough time to read carefully the question. So i'll just leave it here for others to avoid my mistake.
Good luck ya'll

PS: Difficulty is probably under 600, but couldn't edit the tags
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Yeah, upon revising my calculations I got it right, but on my first time I stopped at 4π(9+36) = 180π. Since that's (E), I just selected it. That's the silly mistake that one has to avoid!
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\(\sqrt{45}=\sqrt{9*5}=3\sqrt{5}\)

Hope it's clear.
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