Question -
Answer -
∆PQR is circumscribed by a circle with centre O and radius 8 cm
T is point of contact which divides the line segment OT into two parts such that
QT = 14 cm and TR = 16 cm
Area of ∆PQR = 336 cm²
Let PS = x cm
QT and QS are tangents to the circle from Q
QS = QT = 14 cm
Similarly RU and RT are tangents to the circle
RT = RU = 16 cm
Similarly PS and PU are tangents from P
PS = PU = x
Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm
Now area of ∆PQR = Area of ∆POQ + area of ∆QOR + area of ∆POR