Discussion Forum

1)

Corresponding  to a triangle ABC, match the items given in List -I  with the items given in List II.

The correct match is 

Answer:

Option A

Explanation:

In  $\triangle$ ABC

(A)  We have

      $r r_{2}=r_{1}r_{3}$

$\therefore$    $\frac{\triangle}{s}.\frac{\triangle}{s-b}=\frac{\triangle}{s-a}.\frac{\triangle}{s-c}\Rightarrow\frac{(s-a)(s-c)}{s(s-b)}=1$

$\Rightarrow$      $\tan ^{2} \frac{B}{2}$=1 $\Rightarrow$  $\tan \frac{B}{2}$ = $\tan 45^{0} \Rightarrow$  B=$90^{0}$

$\therefore$      $b^{2}  = a^{2}+c^{2}$

 $\therefore$    A→ II

 (B)   We have

 $r_{1}+r_{2}=r_{3}-r$

$\frac{\triangle}{s-a}+\frac{\triangle}{s-b}=\frac{\triangle}{s-c}-\frac{\triangle}{s}$

  $\Rightarrow$     $\frac{s-b+s-a}{(s-a)(s-b)}=\frac{s-s+c}{s(s-c)}$

 $\Rightarrow$    $\frac{2s-(a+b)}{(s-a)(s-b)}=\frac{c}{s(s-c)}= \tan ^{2} \frac{C}{2}=1$

$\Rightarrow$     $\angle c=90^{0}$

$\Rightarrow$     B  →  III

(C)   $r_{1}=r+2R$

$\frac{\triangle}{s-a}=\frac{\triangle}{s}+\frac{a}{\sin A}\Rightarrow \sin A=\frac{s(s-a)}{\triangle}$

 $\Rightarrow$    $2\sin\frac{A}{2}\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\cot \frac{A}{2}$

$\Rightarrow$  $\sin ^{2} \frac{A}{2}=\frac{1}{2}\Rightarrow \sin \frac{A}{2}=\frac{1}{\sqrt{2}}\Rightarrow \angle A=90^{0}$

$\therefore$    C  →  I