In the given figure, AB is a chord of length 6 cm drawn in a circle with centre 'O'. If OM is the perpendicular bisector of length √27 cm to chord AB then, find the area of segment ABY. (Use π=227 and √27 = 5.2 )
A
6.2 cm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.5 cm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5.8 cm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.25 cm2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 3.25 cm2 As, OM is the perpendicular bisector of AB so, AM = MB = AB2 = 62 = 3 cm ⇒ MB = 3 cm Given: OM = √27 cm
Consider △OMB, Use Pythagoras theorem, OB2 = OM2 + MB2 ⇒OB2 = (√27)2 + 32 ⇒OB2 = 27 + 9 = 36 ⇒ OB = √36 cm ⇒ OB = 6 cm (radius)
In △OMB, sin (∠MOB) = MBOB ⇒ sin (∠MOB) = 36 ⇒sin (∠MOB) = 12 [As, sin 30∘ = 12] ⇒sin (∠MOB) = sin 30∘ ⇒∠MOB = 30∘
so, ∠AOB = 2 ×∠MOB ⇒∠AOB = 60∘
Area of sector = x∘360∘×π×r2
Area of sector OAYB = 60∘360∘×227×62 ⇒ Area of sector OAYB = 18.85 cm2
Area of △OAB = 12×AB×OM = 12×6×5.2 =15.6 cm2
Area of segment ABY = [Area of sector OAYB - Area of △OAB] ⇒ Area of segment ABY = 18.85 - 15.6 = 3.25 cm2