Question

If in $△ABC,\frac{s-a}{\Delta }=\frac{1}{8},\frac{s-b}{\Delta }=\frac{1}{12}=\frac{s-c}{\Delta }=\frac{1}{24}$ then $b$ is equal to

• A. $16$
• B. $20$
• C. $24$
• D. $28$

Answer 1

The correct option is A $16$

As $r_1=\frac{△}{S-a}\Rightarrow r_1=8$, similarly $r_2=12,r_3=24$

Now, as $\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\Rightarrow \frac{1}{r}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}$

$\Rightarrow r=\frac{24}{6}=4$

And $b^2=(r_2-r)(r_3+r_1)=(12-4)(8+24)$

$\Rightarrow b^2=8\times 32=8\times 8\times 4$

$\Rightarrow b=16$