Question

In a $\Delta ABC$ if $r_1=8,r_2=12,r_3=24$ find $a,b,c$

Answer 1

Consider the given

$r_1=8,r_2=12,r_3=24$

Then we know that, In a $\Delta ABC$

$\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$

$\frac{1}{r}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$

$\frac{1}{r}=\frac{3+2+1}{24}$

$\frac{1}{r}=\frac{6}{24}$

$\frac{1}{r}=\frac{1}{4}$

$r=4$

Now, we know that

${\Delta }^2=r{r}_1{r}_2{r}_3$

$\Delta =\sqrt{r{r}_1{r}_2{r}_3}$

$\Delta =\sqrt{4\times 8\times 12\times 24}$

$\Delta =\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 2\times 2\times 2\times 3}$

$\Delta =2\times 2\times 2\times 2\times 2\times 3$

$\Delta =96$

Again, we know that

$r=\frac{\Delta }{s}$

$r=\frac{96}{s}$

$4=\frac{96}{s}$

$s=\frac{96}{4}$

$s=24$

So,

$r_1=\frac{\Delta }{s-a}=\frac{96}{24-a}$

$8=\frac{96}{24-a}$

$24-a=\frac{96}{8}$

$24-a=12$

$a=12$

$r_2=\frac{\Delta }{s-b}=\frac{96}{24-b}$

$12=\frac{96}{24-b}$

$24-b=\frac{96}{12}$

$24-b=8$

$b=16$

$r_3=\frac{\Delta }{s-c}=\frac{96}{24-c}$

$24=\frac{96}{24-c}$

$24-c=\frac{96}{24}$

$24-c=4$

$c=20$

Hence, this is the answer.