A base of a triangle QR touches a circle at S. The other two sides PQ and QR of ∆PQR is extended such a way that they touch the circle at T and U respectively .If the perimeter of ∆PQR is 10 cm, then length of PT is

10 cm

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9 cm

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5 cm

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8 cm

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Solution

The correct option is C
5 cm

We have PT = PU (tangents from a common external points are equal )
QS = QT (tangents from a common external point are equal)
RS = RU (tangents from a common external point are equal)
PQ + QR + PR = 10 cm
2PT = PT + PU = PT + PR +SR
2 PT = PQ + QS + SR + PR
2 PT = PQ + QR + PR
PT = 10/2 = 5 cm

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A triangle PQR is drawn to circumsribe a circle of radius 8 cm such that the segements QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of $Δ P Q R$ is $336 c m^{2}$ , find the sides PQ and PR.