Question

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 ?​

Answer 1

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.

Answer 2

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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