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Properties of Triangle
Properties of Triangle
Medium

Question

Let a, b and c be the length of sides of a triangle ABC such that a+b7=b+c8=c+a9{{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9}. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of Rr{R \over r} is equal to :

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Solution

a+b7=b+c8=c+a9=λ{{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9} = \lambda a+b=7λa + b = 7\lambda b+c=8λb + c = 8\lambda c+a=9λc + a = 9\lambda a+b+c=12λa + b + c = 12\lambda \therefore a=4λ,b=3λ,c=5λa = 4\lambda ,\,b = 3\lambda ,\,c = 5\lambda S=4λ+3λ+5λ2=6λS = {{4\lambda + 3\lambda + 5\lambda } \over 2} = 6\lambda Δ=S(sa)(sb)(sc)=(6λ)(2λ)(3λ)(λ)=6λ2\Delta = \sqrt {S(s - a)(s - b)(s - c)} = \sqrt {(6\lambda )(2\lambda )(3\lambda )(\lambda )} = 6{\lambda ^2} R=abc4Δ=(4λ)(3λ)(5λ)4(6λ2)=52λR = {{abc} \over {4\Delta }} = {{(4\lambda )(3\lambda )(5\lambda )} \over {4(6{\lambda ^2})}} = {5 \over 2}\lambda r=Δs=6λ26λ=λr = {\Delta \over s} = {{6{\lambda ^2}} \over {6\lambda }} = \lambda Rr=52λλ=52{R \over r} = {{{5 \over 2}\lambda } \over \lambda } = {5 \over 2}

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