Mathematics - Properties of Triangle Question with Solution | TestHub

The identity $r_a+r_b+r_c=4R+r$ is always true. Given $r_a+r_b+r_c=9r$, we have $4R+r=9r$, which implies $4R=8r$, or $R=2r$. This is a known condition for a right-angled triangle. Let $C=90^{\circ }$. A. $r=\frac{a+b-c}{2}$ is an identity for any triangle, so $a+b=c+2r$ is correct. B. For a right-angled triangle ($C=90^{\circ }$), $r_a+r_b=\frac{\Delta }{s-a}+\frac{\Delta }{s-b}=\Delta \frac{2s-(a+b)}{(s-a)(s-b)}=\Delta \frac{c}{(s-a)(s-b)}$. Using ${\Delta }^2=s(s-a)(s-b)(s-c)$ and $s-c=r$ for $C=90^{\circ }$, we get $r_a+r_b=\frac{\Delta csr}{{\Delta }^2}=\frac{csr}{rs}=c$. So B is correct. C. From B, $r_a+r_b=c$. We know $\Delta =\frac{1}{2}ab$. Also $r_a{r}_b=s(s-c)=sr$. Since $\Delta =rs$, we have $r_a{r}_b=\Delta$. So $\Delta =\frac{r_a{r}_b}{c}=\frac{r_a{r}_b}{r_a+r_b}$. So C is correct. D. For $C=90^{\circ }$, $r_c=r$. So $r_a{r}_b{r}_c=r_a{r}_{b}r$. We found $r_a{r}_b=sr$. Thus $r_a{r}_b{r}_c=s{r}^2$. For this to be $r^3$, we need $s=r$, which is impossible for a non-degenerate triangle. So D is incorrect.