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. (\(\pi = 3\))}
\(M\) 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. (\(\pi = 3\))}
Show Hint
For perpendicular chords, the shaded right triangle area is straightforward to compute from chord lengths.
Updated On: Mar 20, 2026
- 100 sq. units
- 114.4 sq. units
- 50 sq. units
- 200 sq. units
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The Correct Option is C
Solution and Explanation
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:
\[
\frac{1}{2} \times 10\sqrt{2} \times 5\sqrt{2} = 50 \ \text{sq. units}.
\]
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