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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
As we know tangents drawn from an external point to a circle are equal.
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
So for △PQR
a = PQ = x + 12
b = PR = x + 9
c = QR = 21 cm
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.
PQ = x + 12 = 10.5 + 12 = 22.5 cm