2. In the given figure, a square is inscribed in a
circle with centre O. Find :
A
D
(i) angleBOC
(ii) ZOCB
(iii) ZCOD
(iv) ZBOD
B.
с
Is BD a diameter of the circle ?
Answers
In the given figure we can extend the straight line OB to BD and CO to CA
Then we get the diagonal of the square which intersect each other at 90 degree
by the property of Square.
From the above statement we can see that
∠COD=90°
1) ∠BOC+∠OCD=∠BOD=180°
∠BOC+90° =180°
∠BOC=180°−90°
∠BOC=90°
in △OCB,
∠OBC=∠OCB as they are opposite angles to the two equal sides of an isosceles triangle.
so,
∠OBC+∠OCB+∠BOC=180°
∠OBC+∠OBC+90°−180°
as, ∠OBC=∠OCB
2∠OBC=180°−90°
2∠OBC=90°
∠OBC=45°
as
∠OBC=∠OCB
So,
∠OBC=∠OCB=45°
Yes,
BD is the diameter of the circle.
Given : a square is inscribed in a circle with centre O
To Find : ∠BOC ,∠OCB, ∠COD, ∠BOD
Solution:
AB = BC = BC = AD ( equal sides of Square)
Angle subtended by equal chord are equal
=> ∠AOB = ∠BOC = ∠COD, ∠AOD
∠AOB + ∠BOC + ∠COD + ∠AOD = 360°
=> ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
∠BOC = ∠COD = 90°
∠BOD = ∠BOC+ ∠COD = 90° + 90° = 180°
=> BD is diameter as center lies on BD
in ΔBOC
∠OBC = ∠OCB as OB = OC = radius
=> ∠OBC + ∠OCB + ∠BOC = 180°
=> ∠OBC + ∠OCB + 90° = 180°
=> ∠OBC + ∠OCB = 90°
∠OBC = ∠OCB
=> ∠OCB = 45°
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