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Question
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.
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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