Q. 214.0( 1 Vote )

In Figure 5

Answer :


We know that QT = QS = 12 cm and TR = RU = 9 cm by


the property of tangent of circle.


Let, PS = PU = x cm


Area of ∆ QOR + Area of ∆ POR + Area of ∆ QOP =


Area of ∆ PQR




42 + 2x = 63


x = 21/2 = 10.5 cm


PQ = PS + SQ = 10.5 + 12 = 22.5 cm


PR = PU + UR = 10.5 + 9 = 19.5 cm


So, lengths of side PQ and PR is 22.5 cm and 19.5 cm


respectively.


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