Q. 44.6( 30 Votes )

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.


Answer :

Let AD = x cm, BE = y cm and CF = z cm


As we know that,


Tangents from an external point to a circle are equal,


In given Figure we have


AD = AF = x [Tangents from point A]


BD = BE = y [Tangents from point B]


CF = CE = z [Tangents from point C]


Now, Given: AB = 12 cm


AD + BD = 12


x + y = 12


y = 12 – x…. [1]


and BC = 8 cm


BE + EC = 8


y + z = 8


12 - x + z = 8 [From 1]


z = x – 4…. [2]


and


AC = 10 cm


AF + CF = 10


x + z = 10 [From 2]


x + x - 4 = 10


2x = 14


x = 7 cm


Putting value of x in [1] and [2]


y = 12 - 7 = 5 cm


z = 7 - 4 = 3 cm


So, we have AD = 7 cm, BE = 5 cm and CF = 3 cm


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