Q. 214.0( 1 Vote )

# In Figure 5

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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