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

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