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 the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and the line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?
Note: The figure shown is representative
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