Mathematics
In the given figure, a square is inscribed in a circle with center O. Find :
(i) ∠BOC
(ii) ∠OCB
(iii) ∠COD
(iv) ∠BOD
Is BD a diameter of the circle?
Circles
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Answer
Join OA and OD.
(i) We know that,
Diagonals of a square bisect each other at 90°.
∴ ∠BOC = 90°.
Hence, ∠BOC = 90°.
(ii) From figure,
OB and OC are the radius of the circle.
In △ OBC,
⇒ OB = OC
⇒ ∠OCB = ∠OBC = x (let) [Angle opposite to equal sides are equal]
By angle sum property of triangle,
⇒ ∠OCB + ∠OBC + ∠BOC = 180°
⇒ x + x + 90° = 180°
⇒ 2x + 90° = 180°
⇒ 2x = 180° - 90°
⇒ 2x = 90°
⇒ x =
⇒ x = 45°
⇒ ∠OCB = 45°.
Hence, ∠OCB = 45°.
(iii) We know that,
Diagonals of a square bisect each other at 90°.
∴ ∠COD = 90°.
Hence, ∠COD = 90°.
(iv) From figure,
⇒ ∠BOD = ∠BOC + ∠COD = 90° + 90° = 180°.
Hence, ∠BOD = 180° and BD is the diameter of the circle.
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