Q. 63.8( 8 Votes )

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ΔABC = 21 cm2 then find the lengths of sides AB and AC.


Answer :

Given: ABC that is drawn to circumscribe a circle with radius r = 2 cm and BD = 4 cm DC = 3cm


Also, area(ABC) = 21 cm2


To Find: AB and AC


Now,


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


Then,


FB = BD = 4 cm [Tangents from same external point B]


DC = EC = 3 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 + 4 cm


AC = AE + EC = x + 3 cm


BC = BD + DC = 4 + 3 = 7 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 + 4


b = AC = x + 3


c = BC = 7 cm



And




Squaring both sides,


21(21) = 12x(x + 7)


12x2 + 84x - 441 = 0


4x2 + 28x - 147 = 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 = - 10.5 is not possible as length can't be negative.


So


AB = x + 4 = 3.5 + 4 = 7.5 cm


AC = x + 3 = 3.5 + 3 = 6.5 cm


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