Question

OPQ is the sector of a circle having
centre at 0 and radius 15 cm. If
m<POQ = 30°, find the area enclosed
by arc PQ and chord PQ.​

Answer 1

clearly OP and OQ are radii

OP = OQ = 15 cm

<POQ = 30°

thus

<OPQ = <OQP = (180-30)/2 = 75°

let distance between centre O and chord PQ is OM

in ∆ OMP

sin 75° = OM/OP

sin (45 + 30) = OM/OP

sin 45 cos 30 + cos 45 sin 30 = OM/OP

√3/(2√2) + 1/(2√2) = OM/15

(√3 + 1)/(2√2) = OM/15

OM = 15 (√3 + 1)/(2√2)

again

cos 75° = MP/OP

cos (45 + 30) = MP/OP

cos 45 cos 30 - sin 45 sin 30 = MP/15

√3/(2√2) - 1/(2√2) = MP/15

(√3 - 1)/(2√2) = MP/15

MP = 15 (√3 - 1)/(2√2)

PQ = 2* MP

= 30 (√3 - 1)/(2√2)

area of ∆ OPQ = (1/2)*OM*PQ

= (1/2)*15(√3+1)/(2√2)*30(√3-1)/(2√2)

= 225 (3 - 1)/8

= 225/4 sq cm

area of arc = πr^2 30°/360°

= (22/7)*15^2*(1/6)

= (22/7)*225*(1/6)

= 825/7 sq cm

area between arc and chord

= 825/7 - 225/4

= (825*4 - 225*7)/28

= ( 3300 - 1575 ) / 28

= 1725/28 sq cm