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Question

In the adjoining figure, a triangle 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 the length $$4\,cm$$ and $$3\,cm$$ respectively. if area of $$\triangle ABC$$ is $$21\,{cm}^{2}$$, find the lengths of sides $$AB$$ and $$AC$$.
1400609_6a653f834f52406d98c8e3c2b948eb4f.PNG


Solution

Given :
$$\Delta ABC$$ is circumscribed a circle with centre O and radius $$2 cm.$$
Point D divides $$BC$$ as
$$BD = 4 cm, DC = 3 cm, OD = 2 cm$$
Area of $$\Delta ABC = 21 cm^2$$
Join $$OA, OB, OC, OE$$ and $$OF.$$
From figure:
$$BF$$ and $$BD$$ are tangents to the circle.
So, $$BF = BD = 4 cm$$
$$CD$$ and $$CE$$ are tangents to the circle.
So, $$CE = CD = 3 cm$$
$$AF$$ and $$AE$$ are tangents to the circle.
Let say, $$AE = AF = x cm$$
Now,
Area of $$\Delta ABC= $$ x Perimeter of $$\Delta ABC \times$$ Radius
$$21 = 1/2 [AB + BC + CA] OD$$
$$21 = 1/2 [4 + 3 + 3 + x + x + 4 ) \times 2$$
$$21 = 14 + 2x$$
$$x = 3.5$$
Therefore,
$$AB = AF + FB = 3.5 + 4 = 7.5 cm$$
$$AC = AE + CE = 3.5 + 3 = 6.5 cm$$

1541053_1400609_ans_0fd950d98fbd4ec5ba6002ebd14205a6.png

Mathematics
RS Agarwal
Standard X

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