Question

Assertion :In a $\Delta$ABC, $r_1+r_2+r_3-r=4R$ Reason: $r_1{r}_2+r_2{r}_3+r_3{r}_1={\Delta }^2$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

consider, $r_1+r_2+r_3-r=4R$

we know that $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},r_1\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2},r_2=4R\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2},r_3=4R\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}$

$\Rightarrow 4R[(\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2})+(\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})]=4R$

$\Rightarrow \cos \frac{C}{2}(\sin \frac{A}{2}\cos \frac{B}{2}+\sin \frac{B}{2}\cos \frac{A}{2})+\sin \frac{C}{2}(\cos \frac{A}{2}\cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2})=1$

$\Rightarrow \cos \frac{C}{2}\sin \left(\frac{A+B}{2}\right)+\sin \frac{C}{2}\cos \left(\frac{A+B}{2}\right)=1$

$\Rightarrow \cos \frac{C}{2}\sin \left(\frac{\pi -C}{2}\right)+\sin \frac{C}{2}\cos \left(\frac{\pi -C}{2}\right)=1$

$\Rightarrow {\cos }^2\frac{C}{2}+{\sin }^2\frac{C}{2}=1$

$\Rightarrow 1=1$

Therefore, assertion is correct.
now, $r_1{r}_2+r_2{r}_3+r_3{r}_1={\Delta }^2$

we know that $r_1=\frac{\Delta }{(s-a)},r_2=\frac{\Delta }{(s-b)},r_3=\frac{\Delta }{(s-c)}$

$\Rightarrow {\Delta }^2(\frac{1}{(s-a)(s-b)}+\frac{1}{(s-c)(s-b)}+\frac{1}{(s-a)(s-c)})={\Delta }^2\Rightarrow {\Delta }^2\left(\frac{3s-a-b-c}{(s-a)(s-b)(s-c)}\right)={\Delta }^2$

$\Rightarrow \frac{{\Delta }^2{s}^2}{s(s-a)(s-b)(s-c)}={\Delta }^2$

$\Rightarrow s^2={\Delta }^2$

and reason is incorrect.
Ans: C