ABC is an isosceles triangle inscribed in a circle of radius `rdot` If `A B=A C` and `h` is the altitude from `A` to `B C ,` then triangle `A B C` has perimeter `P=2(sqrt(2h r-h^2)+sqrt(2h r))` and area A= ____________ and = __________ and also `("lim")_(h vec 0) A/(P^3)=______`

Explanation

In `DeltaABC, AB=AC, AD_|_BC(D" is midpoint of "BC)`

Let r= radius of circumcircle
`:." "OA=OB=OC=r`
Now `BD=sqrt(BO^(2)-OD^(2))=sqrt(r^(2)-(h-r)^(2))=sqrt(2rh-h^(2))`
`:." "BC=2sqrt(2rh-h^(2))`
Also` AB^(2)=BD^(2)+AD^(2)=2hr-h^(2)+h^(2)=2hr`.
`:." "AB=AC=sqrt(2hr)`
`:." ""Perimeter",P=2sqrt(2rh-h^(2))+2sqrt(2hr)`
`:." ""Area of "DeltaABC=(1)/(2)xxBCxxAD=hsqrt(2rh-h^(2))`
So,`" "underset(hto0)lim(A)/(P^(3))=(hsqrt(2rh-h^(2)))/(8(sqrt(2rh-h^(2))+sqrt(2hr))^(3))`
`=underset(hto0)lim(h^(3//2)sqrt(2r-h))/(8h^(3//2(sqrt(2r-h)+sqrt(2r))^(3)))`
`=underset(hto0)lim(sqrt(2r-h))/(8[sqrt(2r-h)+sqrt(2r)]^(3))`
`=(sqrt(2r))/(8(sqrt(2r)+sqrt(2r))^(3))=(1)/(128r)`