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

Approach: A perpendicular from the centre bisects each chord, and the radius to an endpoint forms a right triangle with that perpendicular. The \(45^\circ\) angle pins down the first chord; the \(3:2\) ratio gives the second distance, and since the chords sit on opposite sides of a diameter, the figure is a trapezium whose height is the sum of the two distances.

Step 1: Chord \(PQ\) and its distance \(d_1\). Drop \(CM \perp PQ\); then \(M\) is the midpoint and \(\triangle CMQ\) is right-angled at \(M\) with hypotenuse \(CQ = R = 6\sqrt2\) and \(\angle MQC = 45^\circ\). \[ d_1 = CM = R\sin 45^\circ = 6\sqrt2 \cdot \tfrac{1}{\sqrt2} = 6, \qquad MQ = R\cos 45^\circ = 6. \] So \(PQ = 2 \cdot MQ = 12\).

Step 2: Distance \(d_2\) and chord \(SR\). From \(d_1 : d_2 = 3:2\) with \(d_1 = 6\): \[ d_2 = 6 \cdot \tfrac{2}{3} = 4. \] In right \(\triangle CNR\) (with \(CN = d_2 = 4\), \(CR = R\)): \[ NR^2 = R^2 - d_2^2 = 72 - 16 = 56 \implies NR = 2\sqrt{14}, \] so \(SR = 2 \cdot NR = 4\sqrt{14}\).

Step 3: Area of trapezium \(PQRS\). The chords are parallel and on opposite sides of \(C\), so the height between them is \[ h = d_1 + d_2 = 6 + 4 = 10. \] \[ \text{Area} = \tfrac12(PQ + SR)\,h = \tfrac12\big(12 + 4\sqrt{14}\big)\cdot 10 = 5\big(12 + 4\sqrt{14}\big) = 60 + 20\sqrt{14}. \] \[ = 20\big(3 + \sqrt{14}\big). \]

\[ \boxed{20(3 + \sqrt{14}) \text{ sq. cm}} \] (option 1).

Answer 2

Approach: Use the special \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle property directly (equal legs) instead of trigonometric ratios, and the general "chord length from centre distance" relation \(\text{half-chord} = \sqrt{R^2 - d^2}\), to get both chords quickly.

Step 1: Distance to \(PQ\). Drop the perpendicular \(CM\) from centre \(C\) to \(PQ\); \(\triangle CMQ\) is right-angled at \(M\) with \(\angle MQC = 45^\circ\), so it is an isosceles right triangle: \(CM = MQ\). By Pythagoras, \(CM^2 + MQ^2 = R^2 = 72\), and since \(CM=MQ\), \(2\,CM^2 = 72 \Rightarrow CM = 6\). So \(d_1 = 6\) and \(PQ = 2 \times 6 = 12\).

Step 2: Distance to \(SR\) and its length. The ratio \(d_1:d_2 = 3:2\) gives \(d_2 = 4\). Using the general chord relation, half of \(SR\) is \(\sqrt{R^2-d_2^2} = \sqrt{72-16}=\sqrt{56}=2\sqrt{14}\), so \(SR = 4\sqrt{14}\).

Step 3: Area of the trapezium. The chords lie on opposite sides of the diameter, so the height between them is \(d_1+d_2=10\).
\[ \text{Area} = \frac{1}{2}(PQ+SR)\times 10 = \frac{1}{2}(12+4\sqrt{14})\times 10 = 60+20\sqrt{14} = 20(3+\sqrt{14}). \]

\[ \boxed{20(3+\sqrt{14})\text{ sq. cm}} \] option (1).