ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is altitude from A to BC, then $lim h \to 0 \frac{△}{p^{3}}$ is equal to (where $△$ is the area and P is the perimeter of the triangle ABC)

A B C

is an isosceles triangle inscribed in a circle of radius r. If

A B = A C & h

is the altitude from A to BC and P be the perimeter of ABC, then

{lim}_{h \to 0} \frac{Δ}{P^{3}}

equals (where

Δ

is the area of traingle)

Q. 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 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]

and A be the area of the triangle .Find

lim h \to 0 \frac{A}{P^{3}}

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ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude from A to BC, then evaluate limh→0ΔP3, where Δ is area of the triangle and P its perimeter.

The correct option is A 1128rBD=√r2−(h−r)2,AD=h =√2hr−h2AB=√h2+2hr−h2 =√2hrNow limh→0ΔP3=limh→012.h.2√2hr−h2(2√2hr+2√2hr−h2)3=limh→0h32√2r−hh32(2√2r+2√2r−h)3=√2r(2√2r+2√2r)3=1128r

ABC is an isosceles triangle inscribed in a circle of radius r. If $A B = A C$ and h is the altitude from A to BC, then evaluate $l i m h \to 0 \frac{Δ}{P^{3}}$ , where $Δ$ is area of the triangle and P its perimeter.