Mathematics - Properties of Triangle Question with Solution | TestHub

Using the sine rule $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$, the given condition becomes $(2R\sin A{)}^2+(2R\sin B{)}^2+(2R\sin C{)}^2=8R^2$. This simplifies to $4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)=8R^2$, so ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$. This is a known identity for a right-angled triangle (since ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2(1+\cos A\cos B\cos C)$, so $2(1+\cos A\cos B\cos C)=2⟹\cos A\cos B\cos C=0$, implying one angle is $90^{\circ }$). Let $C=90^{\circ }$. A. $r=\frac{a+b-c}{2}$ is an identity for any triangle. So A is correct. B. For a right-angled triangle ($C=90^{\circ }$), $R=c/2$ and $r=(a+b-c)/2$. Thus $R+r=c/2+(a+b-c)/2=(c+a+b-c)/2=(a+b)/2$. So B is correct. C. As derived in the solution for Question 1, for a right-angled triangle, $r_a+r_b=c$, and Area $\Delta =\frac{r_a{r}_b}{c}$. Substituting $c=r_a+r_b$, we get $\Delta =\frac{r_a{r}_b}{r_a+r_b}$. So C is correct. D. The condition $R=2r$ is true only for specific right-angled triangles (e.g., isosceles right triangle, or more generally when $5c^2=8ab$). It is not true for all right-angled triangles (e.g., a 3-4-5 triangle has $R=5/2$ and $r=1$, so $R\ne 2r$). So D is incorrect.