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: √3/2 r2 3√3 r2 √3 r2 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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