## A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. ( = 3.14, √3 = 1.73)

Given central angle = 60˚Radius ,r = 15cmLet chord PQ subtend ∠POQ = 60° at centre.∴ θ = 60°Area of minor segment = r2[(/360) – (sin/2)]= 152[3.14×(60/360) – sin60˚/2]= 225[3.14×(1/6) – √3/4]= 225[(3.14/6) – 1.73/4]= 225[(6.28- 5.19)/12]= 20.44Hence area of minor segment is 20.44cm2.Area of circle = r2= 3.14×152= 3.14×225= 706.5cm2Area of major segment = […]

## In the figure 7.46, if O is the centre of the circle, PQ is a chord. POQ = 90°, area of shaded region is 114 cm2 , find the radius of the circle. ( = 3.14)

Given POQ = = 90°,area of shaded region PRQ = 114cm2Area of segment PRQ = r2[(/360) – (sin/2)]114 = r2[3.14×90/360 – sin90˚/2]114 = r2[3.14×(1/4) – 1/2]114 = r2[0.785-0.5]114 = r2×0.285r2 = 114/0.285 = 400Taking square root on both sidesr = 20cmHence the radius of the circle is 20cm.

## In the figure 7.45, if A is the centre of the circle. PAR = 30°, AP = 7.5, find the area of the segment PQR ( = 3.14)

Given radius AP , r = 7.5Central angle PAR = = 30Area of segment PQR = r2[(/360) – (sin/2)]= 7.52[3.14×(30/360)- sin30˚/2]= 56.25[3.14×(1/12)- 1/4]= 56.25[(3.14/12)- (3/12)]= 56.25×0.14/12= 0.65625cm2Hence area of segment PQR is 0.65625cm2.

## In the figure 7.44, O is the centre of the circle. m(arc PQR) = 60° OP = 10 cm. Find the area of the shaded region. ( = 3.14, √3 = 1.73)

Given radius OP , r = 10cmm(arc PQR) , = 60°Area of segment PQR = r2[(/360) – (sin/2)]= 102[3.14×(60/360)- sin60˚/2]= 100[3.14×(1/6)- √3/4]= 100[(3.14/6)- 1.73/4]= (314/6)-(173/4)= 52.33- 43.25= 9.08cm2Hence area of shaded region is 9.08cm2.

## In figure 7.43, A is the centre of the circle. ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.

Given radius of circle, r = 7√2cmAB = AC [radii of same circle]ABC = ACB = 45° [isosceles triangle theorem]InABC,A = = 90˚ [Angle sum property of triangle]Area of segment BXC = r2[(/360) – (sin/2)]= (7√2)2[3.14×90/360 – (sin90˚)/2]= 98×[(3.14/4)-(1/2)]= 98×[0.785-0.5]= 27.93cm2Hence area of segment BXC is 27.93cm2.