Question

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

	List I		List I
(A)	$r{r}_2=r_1{r}_3$	(I)	$\angle A=90^{\circ }$
(B)	$r_1+r_2=r_3-r$	(II)	$b^2=c^2+a^2$
(C)	$r_1=r+2R$	(III)	$\angle C=90^{\circ }$
		(IV)	$\angle B=120^{\circ }$

The correct match is

• A. A - II , B - III ,C - I
• B. A - II, B - I ,C - III
• C. A - I , B - IV ,C - III
• D. A - III , B - I,C - IV

Answer 1

Answer: (A)

In $\Delta ABC$
(A) We have,
$r{r}_2=r_1{r}_3$
$\therefore \frac{\Delta }{s}\cdot \frac{\Delta }{s-b}=\frac{\Delta }{s-a}\cdot \frac{\Delta }{s-c}\Rightarrow \frac{(s-a)(s-c)}{s(s-b)}=1$
$\Rightarrow {\tan }^2\frac{B}{2}=1$
$\Rightarrow \tan \frac{B}{2}=\tan 45^{\circ }$
$\Rightarrow B=90^{\circ }$
$\therefore b^2=a^2+c^2$
$\therefore A\to II$
(B) We have,
$r_1+r_2=r_3-r$
$\frac{\Delta }{s-a}+\frac{\Delta }{s-b}=\frac{\Delta }{s-c}-\frac{\Delta }{s}$
$\Rightarrow \frac{s-b+s-a}{(s-a)(s-b)}=\frac{s-s+c}{s(s-i)}$
$\Rightarrow \frac{2s-(a+b)}{(s-a)(s-b)}=\frac{c}{s(s-c)}={\tan }^2\frac{C}{2}=1$
$\Rightarrow \angle C=90^{\circ }$
$\therefore B\to III$
(C) $r_1=r+2R$
$\frac{\Delta }{s-a}=\frac{\Delta }{s}+\frac{a}{\sin A}$
$\Rightarrow \sin A=\frac{s(s-a)}{\Delta }$
$\Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\cot \frac{A}{2}$
$\Rightarrow {\sin }^2\frac{A}{2}=\frac{1}{2}$
$\Rightarrow \sin \frac{A}{2}=\frac{1}{\sqrt{2}}\Rightarrow \angle A=90^{\circ }$
$\therefore C\to I$