area of segment question (1 Viewer)

[Kaatie]

MIF 7.4 q10
the length of an arc is 8.9cm and the area of the sector is 24.3cm^2 when an angle of theta is subtended at the centre of the circle. Find the area of the minor segment cut off by theta, correct to one decimal place.

any help greatly appreciated

 

[Timothy.Siu]

area of a sector=1/2 . r^2 theta=24.3
r . theta= 8.9=arc length
subbing that into first equation,
1/2 . r . 8.9=24.3
r=48.6/8.9
theta=(8.9)^2/48.6 (by subbing into 2nd equation)
area of minor segment = area of sector - area of triangle

area of triangle= 1/2 ab sin C = 1/2 . sin ((8.9)^2/48.6) . (48.6/8.9)^2
therefore area of minor segment=9.4cm^2 (1dp)

the values seem pretty dodgey though...

 

[Kaatie]

its correct thanks very much!