Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively.
Let AF = x
Now, in ABC,
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
Again, area of triangle is also equal to the . Therefore,
Area of ΔOBC =
Area of ΔOCA =
Area of ΔOAB =
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
On squaring both sides, we get
Either x+14 = 0 or x − 7 =0
Therefore, x = −14and 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