Question

# 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 $$\Delta ABC = 54 cm^2$$ then find the lengths of sides AB and AC.

Solution

## In the given figure, $$\Delta ABC$$ circumscribed the circle with centre O.Radius $$OD = 3 cm$$$$BD = 6 cm, DC = 9 cm$$Area of $$\Delta ABC = 54 cm^2$$To find :  Length of $$AB$$ and $$AC$$.$$AF$$ and $$EA$$ are tangents to the circle at point A.Let $$AF = EA = x$$$$BD$$ and $$BF$$ are tangents to the circle at point B.$$BD = BF = 6 cm$$$$CD$$ and $$CE$$ are tangents to the circle at point C.$$CD = CE = 9 cm$$Now, new sides of the triangle are:$$AB = AF + FB = x + 6 cm$$$$AC = AE + EC = x + 9 cm$$$$BC = BD + DC = 6 + 9 = 15 cm$$Now, using Heron's formula:Area of triangle $$ABC = \sqrt{s(s - a) (s - b) (s-c)}$$Where $$S = \dfrac{a + b + c}{2}$$$$S = 1/2(x + 6 + x + 9+15) = x + 15$$Area of $$ABC = \sqrt{(x + 15) (x + 15 - (x + 6)) (x + 15 - (x - 9)) (x + 15 - 15)}$$ Or$$54 = \sqrt{(x + 15) (9)(6) (x)}$$Squaring both sides, we have$$54^2 = 54 x (x + 15)$$$$x^2 + 15x - 54 = 0$$Solve this quadratic equation and find the value of x.$$x^2 + 18x - 3x - 54 = 0$$$$x(x + 18) - 3(x + 18) = 0$$$$(x - 3) (x + 18) = 0$$Either $$x = 3$$ or $$x = - 18$$But x cannot be negative.So, $$x = 3$$Answer :-$$AB = x + 6 = 3 + 6 = 9 cm$$$$AC = x + 9 = 3 + 9 = 12 cm$$MathematicsRS AgarwalStandard X

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