Explanation:

Let’s break down what’s going on here:

- Point P is 10 cm from the center O of the circle.

- The circle has a radius of 6 cm.

- Tangents PQ and PR are drawn from P, touching the circle at Q and R.

- We’re asked for the area of the quadrilateral PQOR.

Here’s how to think about it:

- OP = 10 cm, radius = 6 cm, so PQ (and PR) are tangents from a point outside the circle.

- The length of a tangent from a point outside is \( \sqrt{OP^2 - r^2} = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 \) cm.

- OP is the diagonal, QR isn’t (it’s not drawn in the quadrilateral).

- PQOR is actually a kite: OP (10 cm) and OR (6 cm) and PQ (8 cm), and the lines from O to Q and O to R are both radii (6 cm).

- Area of PQOR can be found as the sum of two congruent triangles (OQP and ORP).

Let’s calculate that area:

For triangle OQP:

- OQ = 6 cm, PQ = 8 cm, OP = 10 cm.

- Use Heron’s formula:

- Semi-perimeter, s = (6+8+10)/2 = 12.

- Area = \( \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12×6×4×2} = \sqrt{576} = 24 \) sq. cm.

There are two such triangles (OQP and ORP).

- So, total area = 24 + 24 = 48 sq. cm.

Here’s what that means for your options:

- Option 1: 30 sq.cm Too low.

- Option 2: 40 sq.cm Still not right.

- Option 3: 24 sq.cm That’s just one triangle.

- Option 4: 48 sq.cm This is spot on.

So, your answer is option 4 – 48 sq.cm.