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

Radius of the circle: \[ R = 6\sqrt{2} \text{ cm}. \] Let $d_1$ and $d_2$ be the perpendicular distances from $C$ to chords $PQ$ and $SR$ respectively, with \[ d_1 : d_2 = 3 : 2. \] Step 1: Length of chord $PQ$. Drop a perpendicular from $C$ to $PQ$ at $M$. Then $CM = d_1$ and $M$ is the midpoint of $PQ$. Given $\angle PQC = 45^\circ$. Since $CM \perp PQ$ and $C$ lies on the perpendicular from the center to the chord, $\triangle CMQ$ is right-angled at $M$, with $CQ$ as the hypotenuse. In $\triangle CMQ$: \[ CQ = R = 6\sqrt{2}, \quad \angle MQC = 45^\circ. \] Then \[ CM = CQ \sin 45^\circ = 6\sqrt2 \cdot \frac{1}{\sqrt2} = 6. \] So, \[ d_1 = CM = 6 \text{ cm}. \] Also, \[ MQ = CQ \cos 45^\circ = 6\sqrt2 \cdot \frac{1}{\sqrt2} = 6. \] Since $M$ is the midpoint, \[ PQ = 2\cdot MQ = 12 \text{ cm}. \]
Step 2: Length of chord $SR$. Let $CN = d_2$ be the perpendicular distance from $C$ to chord $SR$, with $N$ the midpoint of $SR$. Given \[ d_1 : d_2 = 3 : 2,\quad d_1 = 6, \] so \[ \frac{6}{d_2} = \frac{3}{2} \Rightarrow 3d_2 = 12 \Rightarrow d_2 = 4 \text{ cm}. \] Now in right-angled $\triangle CNR$ (with $CR = R$ and $CN = 4$): \[ CR^2 = CN^2 + NR^2 \Rightarrow (6\sqrt2)^2 = 4^2 + NR^2 \Rightarrow 72 = 16 + NR^2 \Rightarrow NR^2 = 56 \Rightarrow NR = \sqrt{56} = 2\sqrt{14}. \] Thus, \[ SR = 2\cdot NR = 4\sqrt{14} \text{ cm}. \]
Step 3: Area of quadrilateral $PQRS$. Since $PQ \parallel SR$ and they lie on opposite sides of the center, $PQRS$ is a trapezium. Its height is the distance between the two chords, equal to: \[ h = d_1 + d_2 = 6 + 4 = 10 \text{ cm}. \] Parallel sides: \[ a = PQ = 12,\quad b = SR = 4\sqrt{14}. \] Area formula for a trapezium: \[ \text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(12 + 4\sqrt{14}) \cdot 10 = 5(12 + 4\sqrt{14}) = 60 + 20\sqrt{14}. \] Factor: \[ \text{Area} = 20(3 + \sqrt{14}) \text{ sq. cm}. \] \[ \boxed{20(3 + \sqrt{14})} \]