Construct a trapezium PQRS in which PQ || RS, PQ = 8 cm, QR = 6 cm, RS = 4 cm, angle Q = 60°.

Dear student
Not able to get your answer since you have not mentioned the steps of construction,so follow the proper answer with steps of construction

$$\begin{gathered} \mathrm{In} \mathrm{a} \mathrm{trapezoid} \mathrm{PQRS}, \mathrm{PQ}\parallel \mathrm{SR} \\ \mathrm{So}, \angle \mathrm{Q}+\angle \mathrm{R}=180° \\ \Rightarrow 60+\angle \mathrm{R}=180° \\ \Rightarrow \angle \mathrm{R}=180°-60° \\ \Rightarrow \angle \mathrm{R}=120° \\ \mathrm{Steps} \mathrm{of} \mathrm{construction}: \\ \mathrm{Step} 1: \mathrm{Draw} \mathrm{PQ}=8\mathrm{cm} \\ \mathrm{Step} 2: \mathrm{At} \mathrm{Q}, \mathrm{draw} \angle \mathrm{PQX}=60° \\ \mathrm{Step} 3: \mathrm{From} \mathrm{ray} \mathrm{QX}, \mathrm{cut} \mathrm{off} \mathrm{QR}=6 \mathrm{cm} \\ \mathrm{Step} 4: \mathrm{At} \mathrm{R}, \mathrm{draw} \angle \mathrm{QRY}=120° \\ \mathrm{Step} 5: \mathrm{From} \mathrm{ray} \mathrm{RY}, \mathrm{cut} \mathrm{off} \mathrm{RS}=4 \mathrm{cm} \\ \mathrm{Step} 6: \mathrm{Join} \mathrm{SP} \\ \mathrm{PQRS} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{trapezoid} \end{gathered}$$
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