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In Δ\DeltaABC, If ∠\angleB = 90° then

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∵\because r = (s - b)
tanB2=2(s−b)2=c+a−b2tan\frac{B}{2} = \frac{2(s - b)}{2} = \frac{c + a - b}{2}
∴\therefore R=b2R = \frac{b}{2}
∵\because r=(s−b)tanB2=r = (s - b)tan\frac{B}{2} = ⇒\Rightarrow r = s - b
⇒\Rightarrow r = s - 2R

Source: Grass-G / M-24 Mathematics Paper 31 Dec.docx

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