goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

Actually i had thought of some method like that , just take half the angle between the equal sides as theta and all sides can be expressed in terms of theta and 'r'.

But that method was complicated , so I wanted to know if taking some other variable would make it easy.

how did you get

for -1 as 2k(pi) ?

I think it is pi.

n³ = n²(n+1)²/4

n= n(n+1)/2

therefore

n³= (

n)²

Find the least perimeter of the isosceles triangle in which a circle of radius r can be inscribed?
I would like to know what should be taken as variable to express the perimeter as a function .

let the roots be x₁and x₂.
HM   = 2 x1x2/x1+x2
x1x2   = product of roots   = c/a
x1+x2 = sum of roots        = -b/a

HM   = -2c/b
        =4

x+a-[x+b] = x+b+a-[x+b]-b
= {x+b} + a - b
And you can proceed in the same way. So I think the answer is right .

Find the least perimeter of the isosceles triangle in which a circle of radius r can be inscribed?