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plss answer fast experts 11. Sides ofa triangle are a,b,c and its inradius is R. Prove that area of this triangle is abc 4R Answer to any 7 of the questions from 12 to 21 (7x4=28 score) Rth 0th

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Note : This question involves concept of higher grade but still providing the solution . Solution : Using sine rule : \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = 2 R Now from the figure BD when joined becomes diameter and angle in a semicircle is 90 ° . \Rightarrow ∠ BCD = 90 ° Also, BO = R and BD = 2 R and ∠ BDC = ∠ A (angles in the same segment are equal .) Now in △ BCD sinD = \frac{BC}{BD} = \frac{a}{2 R} So, \frac{a}{sinA} = 2 R Also, area of △ = \frac{1}{2} ab sinC . . . (1) and \frac{c}{sinC} = 2 R \Rightarrow \frac{c}{2 R} = sinC So, (1) becomes Also, area of △ = \frac{1}{2} ab (\frac{c}{2 R}) = \frac{abc}{4 R} Hence Proved .

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plss answer fast expertsplss answer fast experts 11. Sides ofa triangle are a,b,c and its inradius is R. Prove that area of this triangle is abc 4R Answer to any 7 of the questions from 12 to 21 (7x4=28 score) Rth 0th