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In a $\triangle A B C$, if $a=18 \mathrm{~cm}$ and $b=24 \mathrm{~cm}$ and $c=30 \mathrm{~cm}$ then the value of $r_1, r_2$ and $r_3$ are
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The correct answer is:
$12 \mathrm{~cm}, 18 \mathrm{~cm}, 36 \mathrm{~cm}$
But, $\quad \Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\Delta=216$ sq. units
Then, $r_1=\frac{\Delta}{s-a}=\frac{216}{18}=12 \mathrm{~cm}$
or, $\quad r_2=\frac{\Delta}{s-b}=\frac{216}{12}=18 \mathrm{~cm}$
or, $\quad r_3=\frac{\Delta}{s-c}=\frac{216}{6}=36 \mathrm{~cm}$
so, $r_1, r_2, r_3$ are $12 \mathrm{~cm}, 18 \mathrm{~cm}$, and $36 \mathrm{~cm}$
$\Delta=216$ sq. units
Then, $r_1=\frac{\Delta}{s-a}=\frac{216}{18}=12 \mathrm{~cm}$
or, $\quad r_2=\frac{\Delta}{s-b}=\frac{216}{12}=18 \mathrm{~cm}$
or, $\quad r_3=\frac{\Delta}{s-c}=\frac{216}{6}=36 \mathrm{~cm}$
so, $r_1, r_2, r_3$ are $12 \mathrm{~cm}, 18 \mathrm{~cm}$, and $36 \mathrm{~cm}$
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