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Solution:
AB = 1, BD = √3 , OA = OB = OD = 1
The given circle of radius 1 is also circumcircle of triangle ABD
∴ R = 1 for triangle ABD
∴ a/sinA = 2R
∴ √3/sinA = 2R = 2
∴ sinA = √3/2
∴ A = 60°
and hence C = 120o
Also by cosine rule on triangle ABD,
(√3)² = 1² + x² −2x cos60°
or x² −x −2 = 0
∴ x = 2
Now, area ABCD = triangle ABD + triangle BCD
cd = 1 , ⇒ c²d² = 1
Also by cosine rule on triangle BCD we have
(√3)² = c² + d² − 2cd cos120°
= c² + d² + cd
∴ c² + d² = 2 or cd = 1
∴ c² and d² are the roots of t² − 2t +1= 0
∴ c² = d² = 1
∴ BC = 1 = CD and AD = x = 2
Now area ABCD = triangle ABD + triangle BCD
=3/4 – ½ ( 1.2. sin60 ) + ( ½ .c.d. sin120°)
= 3/4 – /2 + /4 c.d
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