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Question

In the adjoining figure O is the center of the circle. AOD = 120. If the radius of the circle be 'r', then find the sum of the areas of quadrilaterals AODP and OBQC:


  1. √3/2 r2

  2. 3√3 r2

  3. √3 r2

  4. None of these 


Solution

The correct options are
C

√3 r2


D

None of these 



Consider triangle ARO and DRO

AO = DO (radius)

RO is common base

AR = RD (Line joining the center to the chord bisects the chord)

Therefore triangles ARO and DRO are congruent.

 AOR = DOR

As AOD = 120

AOR = 60

 In triangle ARO

AOR + ARO + RAO = 180

60 + 90 + RAO = 180

RAO =  30

cos RAO = ARAO

cos 30 =  ARr

AR = ( 32) r
 Area of triangle APO = 12 × PO × AR

= 12 × r × 32 r = 34r2

 Required area = 4 × Area of triangle APO = 3 r2

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