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If r1, r2, r3...

Question

If $r_{1}, r_{2}, r_{3}$ represent the exradius and r represents the in radius then, $\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} - \frac{1}{r}$ = 0

True

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False

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Solution

The correct option is A True
$r_{1} = \frac{Δ}{s - a} a n d r = \frac{Δ}{s} \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} - \frac{1}{r} = \frac{s - a}{Δ} + \frac{s - b}{Δ} + \frac{s - c}{Δ} - \frac{s}{Δ} = \frac{3 s - (a + b + c)}{Δ} - \frac{s}{Δ} = \frac{3 s - (2 s) - s}{Δ} = \frac{3 s - 3 s}{Δ} = 0$
$∴$ Given statement is true

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Similar questions

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

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

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

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

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

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

A B C

a, b, c

△ A B C

a < b < c

r

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

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r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} - \frac{1}{r} = 0

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