Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    In an equilateral triangle, the in-radius, circum-radius and one of the ex-radii are in the ratio

    A) \[2:3:5\]              

    B) \[1:2:3\]

    C) \[1:3:7\]            

    D)        \[3:7:9\]

    Correct Answer: B

    Solution :

    We have;\[\Delta =\frac{\sqrt{3}}{4}{{a}^{2}},\,\,s=\frac{abc}{4\Delta }\] \[\therefore \]  \[r=\frac{\Delta }{S}=\frac{a}{2\sqrt{3}},\,\,R=\frac{abc}{4\Delta }\]                 \[=\frac{{{a}^{3}}}{\sqrt{3}\cdot {{a}^{2}}}=\frac{a}{\sqrt{3}}\] and        \[{{r}_{1}}=\frac{\Delta }{S-a}=\frac{\sqrt{3}/4\cdot {{a}^{2}}}{a/2}=\frac{\sqrt{3}}{2}\] Hence;  \[r:R:{{r}_{1}}=\frac{a}{2\sqrt{3}}:\frac{a}{\sqrt{3}}:\frac{\sqrt{3}}{2}\cdot a=1:2:3\]


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