Question 3 , Question 4.

Dear Student,

Q 4.



$$\begin{gathered} \mathrm{In} ∆\mathrm{PQM}, \\ {\mathrm{PM}}^2 = {\mathrm{PQ}}^2 + {\mathrm{QM}}^{2 } \left[\mathbf{Pythagoras}\mathbf{ }\mathbf{theorem}\right] ....\left(1\right) \\ \mathrm{In} ∆\mathrm{NQR}, \\ {\mathrm{RN}}^2 = {\mathrm{NQ}}^2 + {\mathrm{QR}}^2 \left[\mathbf{Pythagoras}\mathbf{ }\mathbf{theorem}\right] .....\left(2\right) \\ \mathrm{In} ∆\mathrm{NQM}, \\ {\mathrm{MN}}^2 = {\mathrm{NQ}}^2 + {\mathrm{QM}}^2 \left[\mathbf{Pythagoras}\mathbf{ }\mathbf{theorem}\right] .....\left(3\right) \\ \mathrm{adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ {\mathrm{PM}}^2 + {\mathrm{RN}}^2 = {\mathrm{PQ}}^2 + {\mathrm{QM}}^2 + {\mathrm{NQ}}^2 + {\mathrm{QR}}^2 \\ \Rightarrow {\mathrm{PM}}^2 + {\mathrm{RN}}^2 ={\left(2\mathrm{NQ}\right)}^2 + {\mathrm{QM}}^2 + {\mathrm{NQ}}^2 + {\left(2\mathrm{QM}\right)}^2 \\ \Rightarrow {\mathrm{PM}}^2 + {\mathrm{RN}}^2 = 4{\mathrm{NQ}}^2 + {\mathrm{QM}}^2 + {\mathrm{NQ}}^2 + 4{\mathrm{QM}}^2 \\ \Rightarrow {\mathrm{PM}}^2 + {\mathrm{RN}}^2 = 5 \left[{\mathrm{NQ}}^2 + {\mathrm{QM}}^2\right] \\ \Rightarrow \boxed{{\mathbf{PM}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{RN}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{5}\mathbf{ }{\mathbf{MN}}^{\mathbf{2}}} \left[\mathrm{Using} \left(1\right)\right] \end{gathered}$$

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