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 $$AC$$.

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,$$AB = AF + FB = 3.5 + 4 = 7.5 cm$$$$AC = AE + CE = 3.5 + 3 = 6.5 cm$$MathematicsRS AgarwalStandard X

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