ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC.The triangle ABC has perimeter P=2[√(2hr−h2)+√2hr] and A be the area of the triangle .Find limh→0AP3
Options:
A .  
1r
B .  
164r
C .  
1128r
D .  
12r
Answer: Option C : C In △ABC,AB=AC AD⊥BC(DismidpointofBC) Let r = radius of circumcircle ⇒ OA = OB = OC = r Now BD = √BO2−OD2 =√r2−(h−r)2=√2rh−h2 ⇒BC=2√2rh−h2 ∴ Area of △ABC=12×BC×AD =h√2rh−h2 Also limh→0Ap3=h√2rh−h28(√2rh−h2+√2hr)3 =limh→0h3/2√2r−h8h3/2(√2r−h+√2r)3 =limh→0√2r−h8[√2r−h+√2r]3 =√2r8(√2r+√2r)3=√2r8.8.2r√2r=1128r
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