When \( m \) parallel lines are intersected by \( n \) parallel lines, parallelograms are formed by choosing two lines from the \( m \) lines and two lines from the \( n \) lines. The number of ways to select 2 lines from \( m \) lines is \( \binom{m}{2} \) and the number of ways to select 2 lines from \( n \) lines is \( \binom{n}{2} \).
Thus, the total number of parallelograms formed is: \[ \binom{m}{2} \times \binom{n}{2} = \frac{m(m-1)}{2} \times \frac{n(n-1)}{2} \] Simplifying: \[ \frac{m(m-1) \times n(n-1)}{4} \]
Thus, the correct answer is \( \frac{m \times (m - 1) \times (n - 1)}{4} \).
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 
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

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 \( R \)?
