( i ) Since $\square$PQRS is a cyclic quadrilateral,

$$\begin{gathered} \angle \mathrm{QRT} = \angle \mathrm{SPT} ........( 1 ) ( \mathrm{exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{interior} \mathrm{opposite} \mathrm{angle} ) \\ \mathrm{In} ∆\mathrm{TPS} \mathrm{and} ∆\mathrm{TRQ}, \\ \angle \mathrm{PTS} = \angle \mathrm{RTQ} .........( \mathrm{Common} \mathrm{angle} ) \\ \angle \mathrm{QRT} = \angle \mathrm{SPT} .........( \mathrm{From} ( 1 ) ) \\ \Rightarrow ∆\mathrm{TPS} ~ ∆\mathrm{TRQ} ........( \mathrm{AA} \mathrm{similarity} \mathrm{criterion} ) \end{gathered}$$

$$\begin{gathered} ( \mathrm{ii} ) \mathrm{Since} ∆\mathrm{TPS} ~ ∆\mathrm{TRQ}, \mathrm{implies} \mathrm{that} \mathrm{corresponding} \mathrm{sides} \mathrm{are} \mathrm{proportional} \\ \mathrm{i}.\mathrm{e}., \frac{\mathrm{SP}}{\mathrm{QR}} = \frac{\mathrm{TP}}{\mathrm{TR}} \\ \Rightarrow \frac{\mathrm{SP}}{4} = \frac{18}{6} \\ \Rightarrow \mathrm{SP} = \frac{18 \mathrm{X} 4}{6} \\ \Rightarrow \mathrm{SP} =12 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} ( \mathrm{iii} ) \mathrm{Since} ∆\mathrm{TPS} ~ ∆\mathrm{TRQ}, \\ \frac{\mathrm{Area} (∆\mathrm{TPS} ) }{\mathrm{Area} ( ∆\mathrm{TRQ} )} = \frac{{\mathrm{SP}}^2}{{\mathrm{RQ}}^2} \\ \Rightarrow \frac{27 }{\mathrm{Area} ( ∆\mathrm{TRQ} )} = \frac{{12}^2}{4^2} \\ \Rightarrow \mathrm{Area} ( ∆\mathrm{TRQ} ) = \frac{27 \mathrm{X} 4 \mathrm{X} 4}{12 \mathrm{X} 12} \\ \Rightarrow \mathrm{Area} ( ∆\mathrm{TRQ} ) = 3 {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{Area} ( \square \mathrm{PQRS} ) = \mathrm{Area} ( ∆\mathrm{TPS} ) - \mathrm{Area} ( ∆\mathrm{TRQ} ) \\ = 27 - 3 \\ = 24 {\mathrm{cm}}^2 \end{gathered}$$