Question

OAB is a sector of the circle having centre at O and radius 12 cm. If m∠AOB = 45°, find the difference between the area of sector OAB and triangle AOB.

Answer 1

Here, r = 12 cm

θ = 45°

= `(45 xx pi/180)^"c"`

= `(pi/4)^"c"`

Draw AM ⊥ OB
In ΔOAM,

sin 45° = `"AM"/12`

∴ `1/sqrt(2) = "AM"/12`

∴ AM = `12/sqrt(2) xx sqrt(2)/sqrt(2) = 6sqrt(2)"cm"`

∴ A (sector OAB) – A(ΔAOB)

= `1/2"r"^2theta - 1/2 xx "OB" xx "AM"`

= `1/2 xx (12)^2 xx pi/4 - 1/2 xx 12 xx 6sqrt(2)`

= `1/2 xx 144 xx pi/4 - 36sqrt(2)`

= `18pi - 36sqrt(2)`

= `18(pi - 2sqrt(2))"sq.cm"`.