In fig., a circle is inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the length of AD, BE and CF.
Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM
Given: AB = 12 cm, BC = 8 cm and AC = 10 cm. Let, AD = AF = x cm, BD = BE = y cm and CE = CF = z cm (Tangents drawn from an external point to the circle are equal in length) 2(x + y + z) = AB + BC + AC = AD + DB + BE + EC + AF + FC = 30 cm x + y + z = 15 cm AB = AD + DB = x + y = 12 cm z = CF = 15 - 12 = 3 cm AC = AF + FC = x + z = 10 cm y = BE = 15 - 10 = 5 cm x = AD = x + y + z - z - y = 15 - 3 - 5 = 7 cm
Aswered by | 4th Jun, 2014, 03:23: PM
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