ACBM is a segment of a circle such that CM is the perpendicular bisector of AB.

If ICM 4 cm and IABI= 16 cm, calculate:
a the radius of the circle of which ACBM is a
segment
b the angle that chord AB subtends at the
centre of the circle
c the area of the segment.

a. 8.94 cm  b. 127°  c. 12.37 cm²

Step-by-step explanation:

a. Since CM = 4 cm is the perpendicular bisector of AB = 16 cm and r is the radius off the circle. From Pythagoras' theorem,

r² = (AB/2)² + CM²

r = √[(AB/2)² + CM²]

Substituting the values of the variables into r, we have

r = √[(16/2)² + 4²]

r = √[8² + 4²]

r = 4√[2² + 1²]

r = 4√[4 + 1]

r = 4√5 cm

r = 8.94 cm  

b. We know that the length of a chord L = 2rsin(θ/2) where r is the radius of the circle, and θ is the angle subtended by the chord AB.

Since L = 2rsin(θ/2)    and L = AB = 16 cm,

L/2r = sin(θ/2)

taking sine inverse of both sides, we have

θ/2 = sin⁻¹(L/2r)

multiplying both sides by 2, we have

θ = 2sin⁻¹(L/2r)        

substituting the values of the variables, we have

θ = 2sin⁻¹[16/(2 × 8.94)]

θ = 2sin⁻¹[16/17.88]

θ = 2sin⁻¹[0.8949]

θ = 2 × 63.49°

θ = 126.98°

θ ≅ 127°

c. The area of a segment A is given by

A = (θπ/360 - sinθ)r²/2 where θ is the angle subtended by the segment and r = radius of the circle

since θ ≅ 127° and r = 4√5 cm, substituting these values into A, we have

A = (127°π/360 - sin127°)(4√5)²/2

A = (398.93/360 - 0.7988)40

A = (1.1081 - 0.7988)40

A = 0.3093 × 40

A = 12.372 cm²

A ≅ 12.37 cm²