Q. 64.0( 435 Votes )

A chord of a circ

Answer :


Radius (r) of circle = 15 cm

Area of sector OPRQ = (60/360)×Πr2

= 1/6 × 3.14 × (15)2

= 117.75 cm2

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = ×(Side)2

=  × (15)2

=

= 56.25 × 

= 97.3125 cm2

Area of segment PRQ = Area of sector OPRQ - Area of ΔOPQ

= 117.75 - 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle - Area of segment PRQ

= Π(15)2 – 20.4375

= 3.14 ×225 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm2

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
How to Revise ChemistryHow to Revise ChemistryHow to Revise Chemistry45 mins
Nelson Mandela : Long Walk to FreedomNelson Mandela : Long Walk to FreedomNelson Mandela : Long Walk to Freedom42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses