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 cm^{2} 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 cm^{2}

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)

4x^{2} + 84x - 1323 = 0

As we know roots of a quadratic equation in the form ax^{2} + 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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