Q. 464.5( 4 Votes )

# In figure 4, PQ i

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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