Q. 294.0( 5 Votes )

In Fig. , D

Answer :

Given:


DE = 5 cm


DE DF



Join AE and AF


Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.


Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


Property 3: Sum of all angles of a quadrilateral = 360°.


By property 1,


EF = ED = 5 cm(tangent from E)


And,


AE = AF [radius]


By property 2, AED = 90° and AFD = 90°.


Also,


EDF = 90° [ EDEF]


By property 3,


AED + AFD + EDF + EAF = 360°


90° + 90° + 90° + EAF = 360°


EAF = 360° - (90° + 90° + 90°)


EAF = 360° - 270°


EAF = 90°


All angles are equal and adjacent sides are equal AEDF is a square.


Hence, all sides are equal


AE = AF = ED = EF = 5 cm


Hence, Radius of circle = 5 cm

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