Question

\(M\) is the centre of the circle. \(l(QS) = 10\sqrt{2}\), \(\ell(PR) = \ell(RS)\) and \(PR \perp QS\). Find the area of the shaded region. (\(\pi = 3\))}

• A. 100 sq. units
• B. 114.4 sq. units
• C. 50 sq. units
• D. 200 sq. units

Hint:

For perpendicular chords, the shaded right triangle area is straightforward to compute from chord lengths.

Answer 1

The Correct Option is C

Given \(QS = 10\sqrt{2}\), the half-chord length is \(5\sqrt{2}\). Since \(PR \perp QS\) and \(PR = RS\), the geometry forms two right triangles inside the circle. Using Pythagoras and chord properties, the shaded segment area works out to exactly half the product of the perpendicular and base: \[ \frac{1}{2} \times 10\sqrt{2} \times 5\sqrt{2} = 50 \ \text{sq. units}. \]