Mathematics - Properties of Triangles Question with Solution | TestHub

By sine rule in$PQR$$\frac{p}{\mathrm{s}\mathrm{i}\mathrm{n}P}=\frac{q}{\mathrm{s}\mathrm{i}\mathrm{n}Q}=\frac{r}{\mathrm{s}\mathrm{i}\mathrm{n}R}=2R$$\frac{\sqrt{3}}{\mathrm{s}\mathrm{i}\mathrm{n}P}=\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}Q}=\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}R}=2\left(1\right)$$\mathrm{s}\mathrm{i}\mathrm{n}P=\frac{\sqrt{3}}{2}$and$\mathrm{s}\mathrm{i}\mathrm{n}Q=\mathrm{s}\mathrm{i}\mathrm{n}R=\frac{1}{2}$$P={60}^{°}$or${120}^{°}, Q={30}^{°}$and${150}^{°}$$\Rightarrow$As$PQR$is not a right angle triangle Only possibility is$P={120}^{°},\mathrm{ }Q=R={30}^{°}$$\Rightarrow PQR$is an isosceles triangle, hence,$PE$is also a median and$‘O’$will be centroid. ($∆PQE$and$∆PRE$are congruent and$E$is mid point of$QR$)$\Rightarrow r=q=1, P=\sqrt{3}$Option$\left(A\right):$Length of median from$R$$RS=\frac{1}{2}\sqrt{2p^2+2q^2-r^2}=\frac{1}{2}\sqrt{2{\left(\sqrt{3}\right)}^2+2\left(1\right)-1}=\frac{\sqrt{7}}{2}$Option$\left(B\right):$Now, area of$\Delta SEF=\frac{1}{4}\Delta PQR$...$\left(\mathrm{i}\right)$and area of$\Delta SOE=\frac{1}{3}\Delta SEF$...$\left(\mathrm{ii}\right)$from$\left(\mathrm{i}\right)$and$\left(\mathrm{ii}\right)$$\Delta SOE=\frac{1}{12}\Delta PQR$...$\left(\mathrm{iii}\right)$$\Delta PQR=\frac{1}{2}pq \mathrm{s}\mathrm{i}\mathrm{n}R=\frac{1}{2}\times \sqrt{3}\times 1\times \frac{1}{2}=\frac{\sqrt{3}}{4}$$\Delta SOE=\frac{1}{12}\Delta PQR=\frac{\sqrt{3}}{48}$Option$\left(D\right):$Length of$OE$As$O$is centroid$OE$will be$\frac{1}{3}$of$PE$$PE=\frac{1}{2}\sqrt{2q^2+2r^2-p^2}=\frac{1}{2}\sqrt{2+2-3}$$=\frac{1}{2}$$OE=\frac{1}{3} PE=\frac{1}{6}$Option$\left(C\right):$Radius of incircle$=\frac{\Delta }{S}\mathrm{ } \left(\mathrm{i}.\mathrm{e}.\frac{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}}{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}-\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}\right)$$=\frac{\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(\frac{1+1+\sqrt{3}}{2}\right)}$$=\frac{\sqrt{3}}{2\left(2+\sqrt{3}\right)}$$=\frac{\sqrt{3}}{2}\left(2-\sqrt{3}\right)$