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Question

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
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B
1.5 cm2
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C
5.8 cm2
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D
3.25 cm2
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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 = x360 × π× r2

Area of sector OAYB = 60360 × 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

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