Question

In a circle with center $C$ and radius $6\sqrt{2}$ cm, $PQ$ and $SR$ are two parallel chords separated by one of the diameters. If $\angle PQC = 45^\circ$, and the ratio of the perpendicular distance of $PQ$ and $SR$ from $C$ is $3:2$, then the area, in sq. cm, of the quadrilateral $PQRS$ is:

• A. \(20(3 + \sqrt{14})\)
• B. \(20(3\sqrt{2} + \sqrt{7})\)
• C. \(4(3 + \sqrt{14})\)
• D. \(4(3\sqrt{2} + \sqrt{7})\)

Hint:

For chords in a circle: \begin{itemize} \item The perpendicular from the center to a chord bisects the chord. \item Use right triangles with the radius as hypotenuse to relate distances from the center to chord lengths. \item When two parallel chords lie on opposite sides of the center, the distance between them is the sum of their perpendicular distances from the center. \end{itemize}

Answer 1

The Correct Option is A

Step 1: As before, \(d_1=6, PQ=12\) and \(d_2=4, SR=4\sqrt{14}\), with the two chords \(10\) cm apart (\(d_1+d_2\)). Since \(SR > PQ\), picture the trapezium as a central rectangle of width \(PQ\) topped by two identical right triangles that make up the extra width of \(SR\).
Step 2: Rectangle area \(= PQ \times \text{height} = 12 \times 10 = 120\).
Step 3: The two triangles together span the leftover width \(SR - PQ = 4\sqrt{14}-12\) and the same height \(10\), so their combined area is \(\frac{1}{2}(SR-PQ)\times 10 = 5(4\sqrt{14}-12) = 20\sqrt{14}-60\). Total area \(= 120 + 20\sqrt{14}-60 = 60+20\sqrt{14} = 20(3+\sqrt{14})\).
\[ \boxed{20(3+\sqrt{14})\text{ sq. cm}} \]