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° [∵ ED⊥EF]

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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