Question

A chord PQ of a circle with radius r subtends an angle B at the center. Show that the area of the minor segment PQ=1/2.r^2.(B-sinB) and write down the area of the major segment PQ

Answer 1

Step-by-step explanation:

A chord of a circle subtends an angle of theta at the centre of the circle. The area of minor segment cut off by the chord is one eighth of the area ...

Top answer · 2 votes

Area of minor segment = 18 × pir^2 Area of sector - area of AOB = pir^28 In AOB,OM AB, AOM = MOB = theta2 and AM = MB = AB2 In AOM,sin theta2 = AMOA =

Answer 2

Step-by-step explanation:

Area of minor segment=81×πr2

Area of sector-area of △AOB=8πr2

In △AOB,OM⊥AB,∴∠AOM=∠MOB=2θ

and AM=MB=2AB

In △AOM,sin2θ=OAAM=rAM⇒AM=rsin2θ

cos2θ=OAOM=rOM⇒OM=rcos2θ

∴ar△AOM=21×OM×AM=21×rcos2θ×sin2θ=r2sin2θcos2θ

⇒360°θ×πr2−r2sin2θcos2θ=8πr2

⇒r2sin2θcos2θ+8πr2=360°θ×πr2

⇒8r2=360πθ×r2

⇒8sin2θcos2θ+π= 45πθ