Question

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \(\pi = 3.14\) and \(\sqrt3 = 1.73\))

Answer 1

Radius of the circle (r) = \(15\) cm. The area of sector \(OPRQ\) is calculated as \( \frac{60°}{360°} \times \pi r^2 \). Substituting the values, we get \( \frac{1}{6} \times 3.14 \times (15)^2 = 117.75 \, cm^2 \). In triangle \(∆OPQ\), \(∠OPQ = ∠OQP\) since \(OP = OQ\). The sum of angles in a triangle is \(180°\), so \(∠OPQ + ∠OQP + ∠POQ = 180°\). Given \(∠POQ = 60°\), we have \(2∠OPQ = 180° - 60° = 120°\), which means \(∠OPQ = 60°\). Therefore, \(∆OPQ\) is an equilateral triangle. The area of \(∆OPQ\) is given by \( \frac{\sqrt3 } 4 \times (side)^2 \). Substituting the side length (15 cm), the area is \( \frac{\sqrt3 } 4 \times (15)^2 = \frac{225\sqrt3 } 4 \, cm^2 \). This simplifies to \(56.25 \sqrt3 \, cm^2\), which is approximately \(97.3125 \, cm^2\). The area of the minor segment \(PRQ\) is the area of the sector \(OPRQ\) minus the area of \(∆OPQ\). Area of segment \(PRQ\) = \(117.75 - 97.3125 = 20.4375 \, cm^2\). The area of the major segment \(PSQ\) is the area of the circle minus the area of the minor segment \(PRQ\). Area of major segment \(PSQ\) = Area of circle - Area of segment \(PRQ\) Area of major segment \(PSQ\) = \( \pi (15)^2 - 20.4375 \) Area of major segment \(PSQ\) = \( 3.14 \times 225 - 20.4375 \) Area of major segment \(PSQ\) = \( 706.5 - 20.4375 \) Area of major segment \(PSQ\) = \( 686.0625 \, cm^2 \).

The final answer is \(686.0625 \, cm^2\).