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)

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 & h is the altitude from A to BC and P be the perimeter of ABC then limh→0Δp3 equals (where Δ is the area of the triangle)

The correct option is D 1128rArea will be h√2hr−h2And perimeter is 2[√2hr−h2+√2hr]Thenlimh→0+ΔP3=limh→0+h√2hr−h28[√2hr−h2+√2hr]3=limh→0+h.√h√2r−h8.√h.h[√2r−h+√2r]3=limh→0+√2r−h8[√2r−h+√2r]3=√2r8(2√2r)3=1128rHence, option 'C' is correct.

$A B C$ is an isosceles triangle inscribed in a circle of radius $r$ . If $A B = A C$ & $h$ is the altitude from $A$ to $B C$ and $P$ be the perimeter of $A B C$ then $lim h \to 0 \frac{Δ}{p^{3}}$ equals (where $Δ$ is the area of the triangle)