In a $Δ A B C$ if $a = 18, b = 24, c = 30$ and if $l = r_{2} - r_{1}, m = r_{3} - r_{2}$ and $n = r_{3} - r_{1}$
then the descending order of $l, m, n$ is?

n, m, l

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l, m, n

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l, n, m

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n, l, m

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Solution

The correct option is A $n, m, l$
we know that , $s = \frac{18 + 24 + 30}{2} = 36$
and $△ = \sqrt{36 (36 - 18) (36 - 24) (36 - 30)} = 6 * 3 * 12 = 216$

From above Formula,
$r_{1} = \frac{216}{36 - 18} = \frac{216}{18} = 12 r_{2} = \frac{216}{36 - 24} = \frac{216}{12} = 18 r_{3} = \frac{216}{36 - 30} = \frac{216}{6} = 36$

Now, $l = 18 - 12 = 6, m = 36 - 18 = 18, n = 36 - 12 = 24$

so, $n > m > l$

Therefore , Answer is $A$

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