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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
Semiperimeter=s=(28+2x)2=14+x
Area of ΔABC=s(sa)(sb)(sc)
={14+x}{(14+x)14}{(14+x)(6+x)}{(14+x)(8+x)}
=(14+x)(x)(8)(6)
=43(14xx2)
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
43(14x+x2)=28+12+2x+16x+2x
43(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+14x196=0
x2+7x98=0
x2+14x7x98=0
x(x+14)7(x+14)=0
(x+14)(x7)=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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