Square PQRS is inscribed in a circle with centre O. Point T is on the circle as shown in the figure. The side of the square is 9 cm. Then, find the value of  \({\rm{P}}{{\rm{T}}^2} + {\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2}\)?

In ΔPTR ⇒ \({\rm{P}}{{\rm{R}}^2} = {\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2}\) (Diagonal makes 90° at circle)

In ΔSPR ⇒ \({\rm{P}}{{\rm{R}}^2} = {\rm{P}}{{\rm{S}}^2} + {\rm{S}}{{\rm{R}}^2}\)

Thus, \({\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2} = {\rm{P}}{{\rm{S}}^2} + {\rm{S}}{{\rm{R}}^2}\)       ---- (1)

In ΔSTQ ⇒ \({\rm{S}}{{\rm{Q}}^2} = {\rm{S}}{{\rm{T}}^2} + {\rm{T}}{{\rm{Q}}^2}\)

In ΔSQR ⇒ \({\rm{S}}{{\rm{Q}}^2} = {\rm{S}}{{\rm{R}}^2} + {\rm{Q}}{{\rm{R}}^2}\)

Thus, \({\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} = {\rm{S}}{{\rm{R}}^2} + {\rm{Q}}{{\rm{R}}^2}\)        ---- (2)

Add both equation, we have –

\({\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2} + {\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} = {\rm{P}}{{\rm{S}}^2} + {\rm{S}}{{\rm{R}}^2} + {\rm{S}}{{\rm{R}}^2} + {\rm{Q}}{{\rm{R}}^2}\)

⇒ \({\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2} + {\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} = 4{\left( {{\rm{side}}} \right)^2}\)

⇒ \({\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2} + {\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} = 4{\left( 9 \right)^2}\)

⇒ \({\rm{P}}{{\rm{T}}^2} + {\rm{R}}{{\rm{T}}^2} + {\rm{S}}{{\rm{T}}^2} + {\rm{Q}}{{\rm{T}}^2} = 324{\rm{\;}}\)