In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of ΔPQR = 189 cm2 then the length of side PQ is


Given : △PQR that is drawn to circumscribe a circle with radius r = 6 cm and QT = 12 cm QR = 9cm


Also, area(△PQR) = 189 cm2


Let tangents PR and PQ touch the circle at X and Y respectively.


To Find : PQ and QR


Now,


As we know tangents drawn from an external point to a circle are equal.


Then,


QT = QY = 12 cm


[Tangents from same external point B]


TR = RX = 9 cm


[Tangents from same external point C]


PX = PY = x (let)


[Tangents from same external point A]


Using the above data we get


PQ = PY + QT = x + 12 cm


PR = PC + RX = x + 9 cm


QR = QT + TR = 12 + 9 = 21 cm


Now we have heron's formula for area of triangles if its three sides a, b and c are given



Where,



So for △PQR


a = PQ = x + 12


b = PR = x + 9


c = QR = 21 cm



And




Squaring both side


189(189) = 108(x + 21)


7(189) = 4(x + 21)


4x2 + 84x - 1323 = 0


As we know roots of a quadratic equation in the form ax2 + bx + c = 0 are,



So, roots of this equation are,






but x = - 31.5 is not possible as length can't be negative.


So


PQ = x + 12 = 10.5 + 12 = 22.5 cm

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