Question

From the information given in the figure, show that PM=PN=$\sqrt{3}.a$ , where QR= a.

Answer 1

$$\begin{gathered} \mathrm{To}\mathrm{prove}:\mathrm{PM}=\mathrm{PN}=\sqrt{3}a,\mathrm{where}\mathrm{QR}=a. \\ \mathrm{In}△\mathrm{PQR}, \\ \mathrm{PQ}=\mathrm{PR}=\mathrm{QR}=a \\ \mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}. \\ \mathrm{In}\mathrm{an}\mathrm{equilateral}\mathrm{triangle},\mathrm{the}\mathrm{altitude}\mathrm{bisects}\mathrm{the}\mathrm{base}. \\ \therefore \mathrm{QS}=\mathrm{SR}=\frac{a}{2} \\ \mathrm{In}\mathrm{acute}△\mathrm{PMR},\angle \mathrm{M}<90°\mathrm{and}\mathrm{seg}\mathrm{PS}\perp \mathrm{side}\mathrm{MR}. \\ {\mathrm{PR}}^2={\mathrm{PM}}^2+{\mathrm{MR}}^2-2\mathrm{MR}.\mathrm{MS}[\mathrm{By}\mathrm{the}\mathrm{application}\mathrm{of}\text{P}\mathrm{ythagor}a\text{s'}\text{T}\mathrm{heorem}] \\ \Rightarrow a^2={\mathrm{PM}}^2+(\mathrm{MQ}+\mathrm{QR}{)}^2-2\times (\mathrm{MQ}+\mathrm{QR})\times \left(\mathrm{MQ}+\mathrm{QS}\right) \\ \Rightarrow a^2={\mathrm{PM}}^2+(2a{)}^2-2\times \left(2a\right)\times \left(a+\frac{a}{2}\right) \\ \Rightarrow a^2={\mathrm{PM}}^2+4a^2-6a^2 \\ \Rightarrow 3a^2={\mathrm{PM}}^2 \\ \therefore \mathrm{PM}=\sqrt{3}a \\ \mathrm{In}\mathrm{acute}△\mathrm{PNQ},\angle \mathrm{N}<90°\mathrm{and}\mathrm{seg}\mathrm{PS}\perp \mathrm{side}\mathrm{NQ}. \\ {\mathrm{PQ}}^2={\mathrm{PN}}^2+{\mathrm{NQ}}^2-2\mathrm{NQ}.\mathrm{NS}[\mathrm{By}\mathrm{the}\mathrm{application}\mathrm{of}\text{P}\mathrm{ythag}\mathrm{oras}\text{T}\mathrm{heorem}] \\ \Rightarrow a^2={\mathrm{PN}}^2+(\mathrm{NR}+\mathrm{RQ}{)}^2-2\times (\mathrm{NR}+\mathrm{RQ})\times \left(\mathrm{NR}+\mathrm{RS}\right) \\ \Rightarrow a^2={\mathrm{PN}}^2+(2a{)}^2-2\times \left(2a\right)\times \left(a+\frac{a}{2}\right) \\ \Rightarrow a^2={\mathrm{PN}}^2+4{\mathrm{a}}^2-6a^2 \\ \Rightarrow 3a^2={\mathrm{PN}}^2 \\ \therefore \mathrm{PN}=\sqrt{3}a \end{gathered}$$

Thus, PM=PN=$\sqrt{3}.a$

Hence proved.