Q. 124.2( 44 Votes )

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ABC = 54 cm2 then find the lengths of sides AB and AC.


Answer :


Given : ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm


Also, area(ABC) = 54 cm2


To Find : AB and AC


Now,


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


Then,


FB = BD = 6 cm [Tangents from same external point B]


DC = EC = 9 cm [Tangents from same external point C]


AF = EA = x (let) [Tangents from same external point A]


Using the above data, we get


AB = AF + FB = x + 6 cm


AC = AE + EC = x + 9 cm


BC = BD + DC = 6 + 9 = 15 cm


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



Where,



So for ABC


a = AB = x + 6


b = AC = x + 9


c = BC = 15 cm



And




Squaring both sides, we get,


54(54) = 54x(x + 15)


x2 + 15x - 54 = 0


x2 + 18x - 3x - 54 = 0


x(x + 18) - 3(x + 18) = 0


(x - 3)(x + 18) = 0


x = 3 or - 18


but x = - 18 is not possible as length can't be negative.


So


AB = x + 6 = 3 + 6 = 9 cm


AC = x + 9 = 3 + 9 = 12 cm


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