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avatar+165 

In triangle ABC, AB=13,AC=15 , and BC=14. Let  be the incenter. The incircle of triangle ABC touches sides BC, AC, and AB at D, E, and F, respectively. Find the area of quadrilateral AEIF.

 

 Aug 23, 2019
 #1
avatar+165 
+1

never mind, got it.

 Aug 23, 2019
 #2
avatar+106535 
+1

 

Here's the answer for anyone who is curious......

 

Join AI,  BI , CI  and DI

 

The semi-perimeter, S, of triangle ABC  =  [ 13 + 14 + 15] / 2  =  42/2   = 21

 

Using Heron's formula to find the the area, we have that the area of ABC

 

A  =√ [ S (S-13) (S-14) (S - 15)  ]   =  √ [21 * 8 * 7 * 6 ]  =  √ [ 3*7 * 8 * 7 * 3 * 2]  =

 

7*3√ [  8 *2]  =     21√16  =  21 * 4  = 84

 

Now  we have triangles ABI ,   BCI  and  CAI   with altitudes  FI, DI and EI , respectively

 

And each of these altitudes = the radius of the in-circle  

 

So    the area of ABC  =  S * radius...so we have

 

84 =  21 * radius

 

84 / 21  =  radius  =  4  = altitude of  ABI, BCI  and CAI

 

Let   BD, BF  = x     CD, CE  = y  and   AF, AE  = z

 

So.....  BD + CD  = 14     so  x + y  =14     (1)

And BF + AF = 13       so  x + z = 13   ⇒  -x - z = -13       (2)

And  CE  + AE  = 15    so   y + z = 15        (3)  

 

Add (2) and (3)  and we have that  y - x = 2    add this result  to (1)  and we have that

 

2y  = 16

y = 8  

So  z = 7    and x = 6

 

So  the area of  AEIF  =  area of triangle AFI  + area of triangle AEI

 

But  triangles AFI  and AEI  are  congruent right triangles....so we can call the area of AEIF  =  2 area of triangle AFI

 

area of triangle AFI =  (1/2) ( radius) (  AF)  =  (1/2) (4) ( z)  = (1/2)(4)(7)  =  14

 

So

 

2* area of triangle AFI  =  2 * 14 =  28  = area of AEIF 

 

 

 

cool cool cool

 Aug 24, 2019

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