Answer :\:Length of side QR of ∆PQR is 5.5 cm.\:⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━\:Explanation :\:Given :\:Area of two similar triangles, ∆ABC and ∆PQR are 100 cm² and 121 cm² respectively and length side BC of ∆ABC is 5 cm.\:To FIND :\:Length of QR of ∆PQR?\:Solution :\:We are given that ∆ABC ∼ ∆PQR.We clearly know that, if two triangles are similar the ratio of their area is proportional to square of ratio of their corresponding sides.\:THEREFORE,\begin{gathered}\\ :\implies\:\sf \Bigg\{\dfrac{ar\big(\triangle{ABC}\big)}{ar\big(\triangle{PQR}\big)}\Bigg\} = {\Bigg\{\dfrac{BC}{QR}\Bigg\}}^{2}\end{gathered} :⟹{ ar(△PQR)ar(△ABC) }={ QRBC } 2 \begin{gathered}\\ :\implies\:\sf \dfrac{100}{121} = \dfrac{\big(5\big)^2}{\big(QR\big)^2}\end{gathered} :⟹ 121100 = (QR) 2 (5) 2 \begin{gathered}\\ :\implies\:\sf \dfrac{100}{121} = \dfrac{25}{{QR}^{2}}\end{gathered} :⟹ 121100 = QR 2 25 \:By cross multiplying :\begin{gathered}\\ :\implies\:\sf 100\:\times\:{QR}^{2} = 25\:\times\:121\end{gathered} :⟹100×QR 2 =25×121 \begin{gathered}\\ :\implies\:\sf {QR}^{2} \:\times\: 100 = 3025\end{gathered} :⟹QR 2 ×100=3025 \begin{gathered}\\ :\implies\:\sf {QR}^{2} = \dfrac{3025}{100} \end{gathered} :⟹QR 2 = 1003025 \begin{gathered}\\ :\implies\:\sf QR = \sqrt{\dfrac{3025}{100}} \end{gathered} :⟹QR= 1003025 \begin{gathered}\\ :\implies\:\sf QR = {\cancel{\dfrac{55}{10}}} \end{gathered} :⟹QR= 1055 \begin{gathered}\\ :\implies\:\underline{\boxed{\bf{\purple{QR = 5.5}}}}\:\bigstar\end{gathered} :⟹ QR=5.5 ★ \:Therefore, length of side QR of ∆PQR is 5.5 cm.\:━━━━━━━━━━━━━━━━━━━━━━━━━