Question

A circle is inscribed in a triangle ABC such that it touches the sides BC, CA and AB at D,E and F respectively as shown in the figure. If BC = 5cm and AB + AC = 13cm, then AE equals,

(1)2cm
(2)3cm
(3)4cm
(4)6cm

Please post the calculation also​

Answer 1

Answer:

(3) 4cm

Step-by-step explanation:

Two tangents extended from the same point are equal.

Therefore AF=AE, CE=CD, BD=BF.

Also,

AC = AE + EC

BC = BD + DC

AB = AF + BF              .(i)

Now,

AB+ BC + AC                            = 13 + 5

AE + EC + BD + DC + AF + BF = 18 [through (i)]

2(AE + BD + DC)                    = 18

AE + BD + DC                        = 18/2

AE + 5                                    = 9

AE                                          = 4

Therefore AE is equal to 4cm.

I hope you find the solution helpful :)