If the sides fo a triangle are in A.P. as well as in G.P. Then the value of $\frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}}$

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Solution

The correct option is B
0

$2 b = a + c, b^{2} = a c ∴ {(\frac{a + c}{2})}^{2} = a c \Rightarrow \frac{1}{4} (a - c)^{2} = 0 ∴ a = c 2 b = 2 a ∴ a = b = c r_{1} = \frac{s}{s - a}, r_{2} = \frac{s}{s - b}, r_{3} = \frac{s}{s - c} ∴ r_{1} = r_{2} = r_{3} as a = b = c ∴ \frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}} = 1 - 1 = 0$

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Similar questions

\frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}}

If the sides fo a triangle are in A.P. as well as in G.P. Then the value of $\frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}}$

A B C

(\frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}})

r_{1}, r_{2}

r_{3}

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2})

r

r_{1}, r_{2}, r_{3}

A B C

a, b, c

r_{1}, r_{2}, r_{3}

A B C

r

r_{1} r_{2} r_{3} - r (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})

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