circle is circumcentre of the equilateral triangle.hence R=a/root3.next one is incentre so it is a/2root3........... sum of areas of circles is pi{(a square /3)+(a square /12)+........} it is sum of a GP and we get 4/3 pi R squared[1-(1/4)^n]. as n tends to infinity it becomes4/3 pi R squared. sum of triangles,(root3)/4Xa squared(1+1/4+1/16.........) we get,root3XR squared[1-(1/4)^n] limit is root 3XR squared.hence answer for all is d. i may be wrong because none of my answers are matching with anything other than none of these.correct me if i am wrong but do rate me if i am correct.
Moderation Team
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