Step 1: Identify parallel sides and altitudes.
In a parallelogram, opposite sides are parallel: \(PS \parallel QR\) and \(PQ \parallel RS\).
From the figure, \(PT \perp QR\) with \(PT=4\), and \(PV \perp RS\) with \(PV=5\).
Step 2: Compute area using base \(PS\).
Because \(PS \parallel QR\) and \(P \in PS\), the perpendicular distance from \(PS\) to \(QR\) equals \(PT\).
\[
\text{Area} = (\text{base } PS)\times(\text{height to } PS)=PS \times PT = 7 \times 4 = 28 \text{ cm}^2.
\]
Step 3: Equate the area using base \(RS\).
Using base \(RS\) and its corresponding altitude \(PV\):
\[
\text{Area} = RS \times PV \;\Rightarrow\; 28 = RS \times 5 \;\Rightarrow\; RS = \dfrac{28}{5}\text{ cm}.
\]
Final Answer:
\[
\boxed{\dfrac{28}{5}\ \text{cm}}
\]

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
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