Given an isosceles triangle with lateral side of length b, base angle $α < \frac{π}{4}, O, I$ be the circum and in-centres respectively, R, r the circumradius and radius of the incircle. Then

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$Δ = 2 b^{2} \sin 2 α$

$r = \frac{b \sin 2 α}{2 (1 + \cos α)}$

$R = \frac{1}{2} b cosec α$

$O I = |\frac{b \cos (\frac{3 α}{2})}{2 \sin α \cos (\frac{α}{2})}|$

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} a = B C = 2 b \cos α \\ ∴ s = \frac{1}{2} (a + b + c) = b (1 + \cos α) \\ Δ = \frac{1}{2} A B \cdot B C \sin α = \frac{1}{2} b \cdot 2 b \cos α \sin α = \frac{1}{2} b^{2} \sin 2 α \end{array}$
(a) is not true.
$\begin{matrix} r = \frac{Δ}{s} = \frac{\frac{1}{2} b^{2} \sin 2 α}{b (1 + \cos α)} \\ = \frac{b \sin 2 α}{2 (1 + \cos α)} \end{matrix}$
which is given in (b)
$R = \frac{a b c}{4 Δ} = \frac{2 b^{3} \cos α}{4 \cdot \frac{1}{2} b^{2} \sin 2 α} = \frac{1}{2} b cosec α$
which is given in (c)
OI = Distance between circumcentre and incentre
$\begin{matrix} = R \sqrt{[1 - 8 \sin \frac{1}{2} A \sin \frac{1}{2} B \sin \frac{1}{2} C]} \\ = R \sqrt{[(1 - 8 \sin (\frac{π}{2} - α) \sin \frac{1}{2} α \sin \frac{1}{2} α)]} \\ = R \sqrt{[1 - 8 \cos α \sin^{2} (\frac{α}{2})]} \\ = R \sqrt{1 - 4 \cos α (1 - \cos α)} = R | 2 \cos α - 1 | \\ = R |\frac{(2 \cos α - 1) \cos (\frac{α}{2})}{\cos (\frac{α}{2})}| \\ = \frac{b}{2 \sin α} |\frac{2 \cos α \cos \frac{α}{2} - \cos \frac{α}{2}}{\cos \frac{α}{2}}| \\ = |\frac{b \cos (\frac{3 α}{2})}{2 \sin α \cos (\frac{α}{2})}| which is given in (d). \end{matrix}$