1. **State the problem:** Construct quadrilateral PQRS with sides $PQ=4.5$ cm, $QR=6$ cm, $RS=9$ cm, $PS=6$ cm, and diagonal $QS=9$ cm. Find the measure of angle $\angle QRS$.\n\n2. **Draw segment $QS$ of length 9 cm:** This will be the diagonal connecting points Q and S.\n\n3. **Construct triangle $QRS$:**\n- From point Q, draw an arc with radius 6 cm (length $QR$).\n- From point S, draw an arc with radius 9 cm (length $RS$).\n- The intersection of these arcs is point R.\n\n4. **Construct triangle $PQS$:**\n- From point Q, draw an arc with radius 4.5 cm (length $PQ$).\n- From point S, draw an arc with radius 6 cm (length $PS$).\n- The intersection of these arcs is point P.\n\n5. **Calculate angle $\angle QRS$ using the Law of Cosines in triangle $QRS$: $$QR^2 + RS^2 - QS^2 = 2 \times QR \times RS \times \cos(\angle QRS)$$ Substitute values: $$6^2 + 9^2 - 9^2 = 2 \times 6 \times 9 \times \cos(\angle QRS)$$ Simplify: $$36 + 81 - 81 = 108 \times \cos(\angle QRS)$$ $$36 = 108 \times \cos(\angle QRS)$$ $$\cos(\angle QRS) = \frac{36}{108} = \frac{1}{3}$$\n\n6. **Find $\angle QRS$: $$\angle QRS = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ$$\n\n**Final answer:** The measure of angle $\angle QRS$ is approximately $70.53^\circ$.