Three identical circles of circumference 12 are each tangent to one

Great explanation JerryAtDreamScore, how do you arrive at sectors add up to a half-circle? Thanks

Yet another example of a question where we don't need to know much of the "real" math to get the question right pretty quickly, with a high degree of confidence that we are right, and with low risk of making a silly error.

If the circumference of the circles is 12(pi), the diameter is 12, which makes the radius 6. The area of each circle is 36(pi) One of those wedges is 1/6th of a circle, so each wedge has area 6(pi). The shaded area is what, maybe a little less than half a wedge? Cool, let's call it 2(pi)? That's around 6.

Let's check the answer choices.
\(\sqrt{3}\) is right around 1.7.
(A) The thing we are subtracting is larger than the thing we are subtracting it from...A is negative. Wrong.
(B) The thing we are subtracting is larger than the thing we are subtracting it from...B is negative. Wrong.
(C) The thing we are subtracting is larger than the thing we are subtracting it from...C is negative. Wrong.
(D) 61.2 - 55.8 = 5.4 Keep it.
(E) 61.2 - 18.6 = 42.6 Way too big.

Answer choice D.


ThatDudeKnowsBallparking

We know that triangle ACE is equilateral, so each of its angles is 60 degrees. 60 is one-sixth of 360, so the area of each wedge must be one-sixth of a circle. Make sense?

Thanks ThatDudeKnows and got your point of one-sixth of a circle now. However how did you derive it into a half-circle? Thanks

If one of them is one-sixth of a circle and we have three of them, that means we have 3*(1/6), which equals 1/2.

But I probably wouldn't mess around with all the geometry in the first place!
https://gmatclub.com/forum/three-identi ... l#p3024890

Brilliant and crystal clear now with your absolute straightforward innovative explanation and posts.
Very much appreciated ThatDudeKnows and you rock.