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 (r) of circle = \(15\) cm   
Area of sector \(OPRQ\) = \(\frac{60°}{360°} \times \pi r^2\)
= \(\frac{1}{6} \times 3.14 \times  (15)^2\)
= \(117.75\,cm^2\)
In \(∆OPQ\), 
\(∠OPQ = ∠OQP\)         (As \(OP = OQ\))  
\(∠OPQ + ∠OQP + ∠POQ = 180°\)
\(2∠OPQ = 120°\)
  \(∠OPQ = 60°\) 
\(∆OPQ\) is an equilateral triangle.  
Area of \(∆OPQ\) = \(\frac{\sqrt3 } 4 \times (side)^2\)
= \(\frac{\sqrt3 } 4 \times (15)^2\)

= \(\frac{225\sqrt3 } 4 cm^2\)

= \(56.25 \sqrt3\)
= \(97. 3125 \, cm^2\)
Area of segment \(PRQ\) = Area of sector \(OPRQ\) − Area of \(∆OPQ\) 
                                    = \(117.75 − 97.3125\)  = \(20.4375\) \(cm^2\)
Area of major segment \(PSQ\) = Area of circle − Area of segment \(PRQ\) 
                                              = \(\pi (15)^2  - 20.4375\)
                                              = \(3.14 \times 225 - 20.4375\)
                                              = \(706.5 - 20.4375\)
                                              = \(686. 0625\) \(cm^2\)

So, the answer is \(686. 0625\ cm^2\).