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ABCD is a cyclic quadrilateral in which AB = BC = 8 cm, O is the center of the circle and one of the diagonal of the cyclic quadrilateral passes through the center of the circle. Find the radius of the circle.

ABCD is a cyclic quadrilateral in which AB = BC = 8 cm, O is the center of the circle and one of the diagonal of the cyclic quadrilateral passes through the center of the circle. Find the radius of the circle. Correct Answer 4√2 cm

Given: 

ABCD is a cyclic quadrilateral

AB = BC = 8 cm

'O' is the center of a circle

Concept used:

The angle made in a semi-circle is 90° 

Pythagoras theorem,

hypotenuse2 = base2 + perpendicular2

Calculation:

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Let us assume AC be diagonal that passes through the center

From the above figure, we can see that AC is also the diameter of the circle so by property of circle ∠ABC = 90° 

So, ΔABC becomes a right angle triangle then AC also becomes hypotenuse of the right-angle triangle.

The radius of the circle = AC/2 

hypotenuse2 = 82 + 82

⇒ hypotenuse2 = 64 + 64

⇒ hypotenuse2 = 128

⇒ hypotenuse = √(128)

⇒ hypotenuse = √(2 × 8 × 8)

⇒ hypotenuse = 8√2 cm

The radius of circle = (8√2)/2 

⇒ 4√2 cm

∴ The radius of circle is 4√2 cm.     

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