In a $△ A B C, r r_{2} + r_{1} r_{3} =$

b c

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a c

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a b

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\frac{a + b + c}{3}

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Solution

The correct option is B $a c$
$r r_{2} + r_{1} r_{3} = \frac{Δ}{s} \cdot \frac{Δ}{s - b} + \frac{Δ}{s - a} \cdot \frac{Δ}{s - c}$
$= \frac{Δ^{2} [(s - a) (s - c) + s (s - b)]}{s (s - a) (s - b) (s - c)}$
$= \frac{Δ^{2} [2 s^{2} - s (a + b + c) + a c]}{Δ^{2}}$

$= 2 s^{2} - 2 s^{2} + a c = a c$

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Similar questions

In a $△ A B C, (r_{1} + r_{3}) \sqrt{\frac{r r_{2}}{r_{1} r_{3}}}$ is equivalent to

△ A B C

B

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

Question 10

In $Δ A B C$ ,

a) AB + BC > AC
b) AB + BC < AC
c) AB + AC < BC
d) AC + BC < AB

$I n Δ A B C, i f ∠ C > ∠ B, t h e n (a) B C > A C (b) A B > A C (c) A B < A C (d) B C < A C$

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