Answer :
To Find: AB and AC
Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.
In ∆ABC,
From the theorem which states that the lengths of two tangents drawn from an external point to a circle are equal.So,
CF = CD = 6cm (Tangents on the circle from point C)
BE = BD = 8cm (Tangents on the circle from point B)
AE = AF = x (Tangents on the circle from point A)
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
where, a, b and c are sides of triangle and s is semi perimeter.
2S = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
S =
=√(14 + x)(x)(8)(6)
Also, area of triangle =
And from the given figure it is clear that
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
Area of ΔOBC =
Area of ΔOCA =
Area of ΔOAB =
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
⇒
⇒
⇒
⇒
⇒ 3x = 14 + x
⇒ 2x = 14
⇒ x = 7
Therefore, x = 7
Hence,
AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm
Hence, AB = 15 cm and AC = 13 cm
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