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In the given figure, *BDC* is a tangent to the given circle at point *D* such that *BD* = 30 cm and *CD *= 7 cm. The other tangents *BE* and *CF* are drawn respectively from *B* and *C* to the circle and meet when produced at *A* making BAC a right angle triangle. Calculate (ii) radius of the circle.

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#### Solution

Let us join the point of contact *E* with the centre of the circle say *O*. Also, let us join the point of contact *F* with the centre of the circle *O*. Now we have a quadrilateral *AEOF*.

In this quadrilateral we have,

`∠EAD = 90^0`(Given in the problem)

`∠oda = 90^0`(Since the radius will always be perpendicular to the tangent at the point of contact)

`∠OEA = 90^0`(Since the radius will always be perpendicular to the tangent at the point of contact)

We know that the sum of all angles of a quadrilateral will be equal to `360^o`. Therefore,

`∠EAD + ∠ ODA + ∠ EOD + ∠ OEA = 360^o`

`90^0+90^o +90^o + ∠EOD = 360^o`

` ∠EOD = 90^o`

Since all the angles of the quadrilateral are equal to 90° and the adjacent sides are equal, this quadrilateral is a square. Therefore all the sides are equal. We have found that

*AF* = 5

Therefore,

*OD* = 5

*OD* is nothing but the radius of the circle.

Thus we have found that *AF* = 5 cm and radius of the circle is 5 cm.

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