Diagonals of a trapezium \(PQRS\) intersect at \(O\). If \(PQ\parallel RS\) and \(PQ=3\,RS\), find \(\dfrac{\operatorname{ar}(\triangle POQ)}{\operatorname{ar}(\triangle ROS)}\).
\(9:1\)
Step 1: The trapezium \(PQRS\) has two parallel sides: \(PQ\) and \(RS\). It is also given that \(PQ = 3 \times RS\).
Step 2: The diagonals of the trapezium meet at point \(O\). This divides the trapezium into two pairs of triangles. We are interested in triangles \(\triangle POQ\) and \(\triangle ROS\).
Step 3: These triangles lie between the same pair of diagonals and also between the parallel lines \(PQ\) and \(RS\). Because of this, the triangles \(\triangle POQ\) and \(\triangle ROS\) are similar triangles (their shapes are the same, only their sizes differ).
Step 4: The ratio of sides of these two similar triangles will be the same as the ratio of the parallel sides of the trapezium. So,
\[ \dfrac{PO}{RO} = \dfrac{PQ}{RS} = 3:1 \]
Step 5: When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
So,
\[ \dfrac{\operatorname{ar}(\triangle POQ)}{\operatorname{ar}(\triangle ROS)} = \left(\dfrac{PQ}{RS}\right)^2 = 3^2 = 9:1 \]
Final Answer: The required ratio is \(9:1\).