ABC is an isosceles triangle inscribed in a cirle of radius r. If AB=AC and h is the altitude from A to BC, then the triangleABC has perimeter P = 2(2hr−h2)+2hr and area A then limh→0AP3

In ∆ABC, AB = AC     AD ⊥BC (D is mid point of BC) Let r = radius of circumcircle ⇒OA =OB =OC =r Now BD = BO2−OD2=r2−(h−r)2=2r−h2⇒BC=22rh−h2 ∴  Area  of  △ABC=12×BC×AD=h2rh−h2 Also limh→0Ap3=h2rh−h28(2rh−h2+2hr)3=limh→0h3/22r−h8h3/2(2r−h+2r)3 =limh→02r−h8(2r−h+2r)3 =        2r8(2r+2r)3=2r8.8.2r.2r=1128r