In Fig, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is
A. 3 cm
B. 5 cm
C. 4 cm
D. 6 cm
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Radius = 5 cm
Step-by-step explanation:
DE and DF are tangents from an external point D to a circle with centre A
DE = DF ( tangent from D)
=> DE = DF = 5 cm
∠AED = ∠AFD = 90° ( Tangent)
∠EDF = 90° DE ⊥ DF
=> in Quadrilateral AEDF
all angles are 90°
=> AEDF is a rectangle
=> AE = DF & AF = DE
=> AE = AF = 5 cm
=> AEDF is a square
=> Radius = AE = AF = 5 cm
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