CAT 2025

Slot 1

QA

Question & SolutionQues & Sol

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 $\angle QC=45^{circ}$, 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

1. Concept Used

• Topic: Circle Geometry — Chords, Perpendicular Distance from Center, and Area of a Trapezium
• Formula: $\text{Area of Trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$

Also, for a chord of length $2l$ at perpendicular distance $d$ from the center of a circle with radius $r$: $l^2 + d^2 = r^2$

2. Calculation

The radius of the circle is $r = 6\sqrt{2}$ cm. The two parallel chords PQ and SR lie on opposite sides of a diameter, so the total height of the trapezium PQRS equals the sum of the two perpendicular distances from C.

Step 1: Find the length of chord PQ using the given angle.

We are given $\angle QCR = 45^\circ$ (interpreting this as $\angle PCQ = 90^\circ$ since the triangle PCQ is isosceles with $CP = CQ = 6\sqrt{2}$ and $\angle CQP = \angle CPQ = 45^\circ$, so $\angle PCQ = 90^\circ$).

Using the Pythagorean theorem in triangle PCQ: $PQ^2 = CP^2 + CQ^2 = (6\sqrt{2})^2 + (6\sqrt{2})^2 = 72 + 72 = 144$ $PQ = 12 \text{ cm}$

Step 2: Find the perpendicular distance CA from C to chord PQ.

Let A be the midpoint of PQ, so $AQ = \frac{PQ}{2} = 6$ cm. In right triangle CAQ: $CA^2 + AQ^2 = CQ^2$ $CA^2 + 36 = 72$ $CA^2 = 36 \implies CA = 6 \text{ cm}$

Step 3: Use the ratio of perpendicular distances to find CB.

The ratio of perpendicular distances of PQ and SR from C is $CA : CB = 3 : 2$. Setting $CA = 3x$ and $CB = 2x$: $3x = 6 \implies x = 2$ $CB = 2x = 4 \text{ cm}$

Step 4: Find the length of chord SR.

Let B be the midpoint of SR. In right triangle CBS: $BC^2 + BS^2 = CS^2$ $16 + BS^2 = (6\sqrt{2})^2 = 72$ $BS^2 = 56 \implies BS = 2\sqrt{14} \text{ cm}$ $SR = 2 \times BS = 4\sqrt{14} \text{ cm}$

Step 5: Find the height AB of the trapezium.

Since PQ and SR are on opposite sides of the diameter: $AB = CA + CB = 6 + 4 = 10 \text{ cm}$

Step 6: Calculate the area of trapezium PQRS. $\text{Area} = \frac{1}{2} \times (PQ + SR) \times AB = \frac{1}{2} \times (12 + 4\sqrt{14}) \times 10$ $= 5 \times (12 + 4\sqrt{14}) = 60 + 20\sqrt{14} = 20(3 + \sqrt{14}) \text{ sq. cm}$

3. Solution

Answer = Option C ✅

The area of quadrilateral PQRS is $20(3 + \sqrt{14})$ sq. cm.