In a $\triangle A B C$, let $a, b, c, s, r, R, I, S, r_1, r_2, r_3$ stand for their usual meaning. Then Match the items of List-I with those of the items of List-II.
The correct match is

• Correct answer
  A

  
  
• B

  
  
• C

  
  
• D

  
  

Solution

A. $\because \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{(s-b)(s-c)}{\Delta}$ and $\frac{r}{s-a}=\frac{\Delta}{s(s-a)}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{(s-b)(s-c)}{\Delta}$ $\therefore \tan \frac{A}{2}=\frac{r}{s-a}=\frac{(s-b)(s-c)}{\Delta}$ B. From the figure, $r=\mathrm{IF}=(\mathrm{AI}) \sin \frac{A}{2}=\mathrm{AI} \sqrt{\frac{(s-b)(s-c)}{b c}}$
C. SI = distance between circumcentre and in centre $=\sqrt{R^2-2 r R}$ $\therefore \quad(\mathrm{SI})^2+2 R r=R^2$ D. As we know that, $r_1+r_2+r_3=4 R+r$ So, on squaring both sides, we get $r_1^2+r_2^2+r_3^2=(4 R+r)^2-2\left(r_1 r_2+r_2 r_3+r_3 r_1\right)$ $=(4 R+r)^2-2\left[\frac{\Delta^2}{(s-a)(s-b)}+\frac{\Delta^2}{(s-b)(s-c)}\right.$ $\left.+\frac{\Delta^2}{(s-c)(s-a)}\right]$ $=(4 R+r)^2-2[s(s-c)+s(s-a)+s(s-b)]$ $=(4 R+r)^2-2 s[3 s-2 s]=(4 R+r)^2-2 s^2$ $=(4 R+r+\sqrt{2} s)(4 R+r-\sqrt{2} s)$

Hence, the correct answer is option: 'A'.