1. **State the problem:** We have two congruent polygons RQOP and XY with given angles and side lengths. We need to find the length of side $QR$ in polygon RQOP. 2. **Understand congruence:** Congruent polygons have all corresponding sides and angles equal. Since polygons RQOP and XY are congruent, corresponding sides have equal lengths. 3. **Identify corresponding sides:** Given side $OR = 30$ cm and $PO = 26$ cm in polygon RQOP, their corresponding sides in polygon XY have the same lengths. 4. **Use given angles and sides:** Since the polygons are congruent, side $QR$ corresponds to side $XY$ in polygon XY. However, the length of $XY$ is not given directly. 5. **Apply the Law of Cosines in triangle $QPO$ to find $QR$:** Triangle $QPO$ has sides $QP$, $PO=26$ cm, and angle at $P$ is $68^\circ$ (from the problem's angle data). We want to find $QR$, which corresponds to side $QP$. 6. **Calculate $QR$ using Law of Cosines:** $$QR^2 = QP^2 = PO^2 + RQ^2 - 2 \times PO \times RQ \times \cos(\angle P)$$ But we don't have $RQ$ yet, so instead, use the triangle $ORQ$ with sides $OR=30$ cm, angle $R=87^\circ$, and angle $Q=68^\circ$. 7. **Find angle $O$ in triangle $ORQ$:** $$\angle O = 180^\circ - 87^\circ - 68^\circ = 25^\circ$$ 8. **Use Law of Sines in triangle $ORQ$ to find $QR$:** $$\frac{QR}{\sin(87^\circ)} = \frac{OR}{\sin(25^\circ)}$$ $$QR = \frac{OR \times \sin(87^\circ)}{\sin(25^\circ)} = \frac{30 \times \sin(87^\circ)}{\sin(25^\circ)}$$ 9. **Calculate values:** $$\sin(87^\circ) \approx 0.9986, \quad \sin(25^\circ) \approx 0.4226$$ $$QR = \frac{30 \times 0.9986}{0.4226} \approx \frac{29.958}{0.4226} \approx 70.87$$ 10. **Final answer:** $$\boxed{QR \approx 70.87 \text{ cm}}$$