Two mutually perpendicular chords PQ and RS meet at a point M inside the circle such that PM = 12 cm, SM = 3 cm, QM = 4 cm. Find the area of the circle. (use π

To find the area of the circle, we first need to find the radius.

Given that the chords PQ and RS are perpendicular, we can use the Perpendicular Chord Bisector Theorem which states that the line joining the point of intersection M to the center O of the circle bisects the chords at right angles.

This means that OM is the hypotenuse of two right triangles, △PMO and △SMO.

In △PMO, we have PM = 12 cm and QM = 4 cm. Using the Pythagorean theorem, we can find OM:

OM = √(PM² + QM²) = √(12² + 4²) = √160 = 4√10 cm

In △SMO, we have SM = 3 cm. Since OM bisects RS at right angles, RM = SM = 3 cm.

So, the diameter of the circle is OM + RM = 4√10 cm + 3 cm.

The radius of the circle is half of the diameter, so the radius r = (4√10 cm + 3 cm) / 2 = 2√10 cm + 1.5 cm.

The area of the circle is given by the formula πr².

Substituting the value of r, we get:

Area = π(2√10 + 1.5)² = π(40 + 6√10 + 2.25) = 42.25π + 6π√10 cm².