  Question

# A triangle ABC is drawn to circumscribe a circle of radius 3cm , such that the segments BD and DC are respectively of lengths 6cm and 9cm . If the area of triangle ABC is 54cm2 , find the lengths of sides AB and AC .

Solution

## CF = CD = 6cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C) BE = BD = 8cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B) AE = AF = x (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A) Now, AB = AE + EB = x + 8 Also, BC = BD + DC = 8 + 6 = 14 and CA = CF + FA = 6 + x Now, we get all the sides of a triangle so its area can be find out by using Heron’s formula as: 2s = AB + BC + CA = x + 8 + 14 + 6 + x = 28 + 2x ⇒Semi−perimeter=s=(28+2x)2=14+x Area of ΔABC=√s(s−a)(s−b)(s−c) =√{14+x}{(14+x)−14}{(14+x)−(6+x)}{(14+x)−(8+x)} =√(14+x)(x)(8)(6) =4√3(14x−x2) Again, area of triangle is also equal to the 12×base×height Therefore, Area of ΔOBC 12×OD×BC=12×4×14=28 Area of ΔOCA= 12×OF×AC=12×4×(6+x)=12+2x Area of ΔOAB 12×OE×AB=12×4×(8+x)=16+2x Area of ΔABC=Area of ΔOBC+Area of ΔOCA+Area of ΔOAB 4√3(14x+x2)=28+12+2x+16x+2x ⇒4√3(14x+x2)=56+4x ⇒√3(14x+x2)=14+x On squaring both sides, we get ⇒3(14x+x2)=(14+x)2 ⇒42x+3x2=196+x2+28x ⇒2x2+14x−196=0 ⇒x2+7x−98=0 ⇒x2+14x−7x−98=0 ⇒x(x+14)−7(x+14)=0 ⇒(x+14)(x−7)=0 Either x + 14 = 0 or x – 7 = 0 Therefore, x = - 14 and 7 However, x = -14 is not possible as the length of the sides will be negative. Therefore, x = 7 Hence, AB = x + 8 = 7 + 8 = 15 cm CA = 6 + x = 6 + 7 = 13 cm  Suggest corrections   