In a triangle ABC right-angled at B, the inradius is
A
AB+BC−AC
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B
AB+AC−AC
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C
AB+BC−AC2
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D
none of these.
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Solution
The correct option is DAB+BC−AC2 Let AE=AF=α BD=BE=β CD=CF=γ ∵AB+BC+CA=2(α+β+γ) ⇒2s=2(α+β+γ) ∴α+β+γ=s ⇒α=s−a,β=s−b,γ=s−c In △BID,tanπ4=rβ ⇒1=rβ ⇒β=r=s−b =a+b+c2−b =a+c−b2 =AB+BC−CA2