Answer :

In the given figure, PQ = 16 cm

PO = 10 cm (radius)


TP = TQ [Tangents drawn from an external point to the same circle are equal]


ΔTPQ is an isosceles triangle


OT is the perpendicular bisector of PQ


PR = 8 cm


In ΔTPR,


TP2 = TR2 + PR2 [Pythagoras theorem]


TP2 = TR2 + 82 …. (i)


TPO = 90° [A tangent is perpendicular to the radius at the point of tangency]


In ΔTPO,


TO2 = TP2 + PO2 [Pythagoras theorem]


TO2 = TP2 + 102


TP2 = TO2 – 100 …. (ii)


In ΔOPR,


OR2 = OP2 – PR2 [Pythagoras theorem]


OR2 = 102 – 82


OR = √36 = 6 cm


Equation eq (i) and eq (ii),


TR2 + 64 = TO2 – 100


TR2 + 64 = (TR + OR)2 – 100 [From the figure, (TR + OR) = TO]


TR2 + 64 = TR2 + 36 + 12TR – 100


12TR = 128


TR =


Substituting the value of TR in eq (i), we get


TP2 =


TP2 =


TP =


TP = 13.33 cm


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