A circle inscribed in a triangle ABC touches its sides AB, BC and AC at the points D, E and F, respectively. If AB = 18 cm, BC = 15 cm and AC = 13 cm, then the value of AD + BE + CF is:

Solution (By Examveda Team)

2(x + y + z) = 46
x + y + z = 23
AD + BE + CF = 23
Note:-

Perimeter of Δ = 2(x + y + z)
Then, AD + BE + CF = x + y + z
If a circle is inscribe into a triangle touches the side AB. BC & CA at D, E & F.
Then AD + BE + CF = $$\frac{1}{2}$$ × Perimeter of ΔABC