help

We  can find  the height  of  this triangle  as  follows

Using  Heron's Formula  we have  the area as

sqrt  (9 (9-8) (9-6) (9 - 4))  = sqrt  ( 9 * 3 * 5)  =  3sqrt(15)

So.....to find the height we  have

Area  = (1/2)(BC) height

3sqrt (15)  =  (1/2)(4) height

3sqrt (15)  = 2 * height

(3/2)sqrt (15)  = 1.5 sqrt (15)  =  height  = y coordinate of A

And we  can find  the  x coordinate  of A  by the Pythagorean Theorem

sqrt  [ AC^2  -  height of ABC^2 ]   = sqrt  (8^2   -(1.5 sqrt (15))^2  )  = sqrt (30.25)  =  5.5

So....the coordinates  of A  = (5.5, 1.5sqrt (15))

We  can use a formula to find  the coordinates  of  the center  of the incircle

Let B = (4,0)  and C  = (4,0)

x coordinate  of incenter  =

[ Ax* a  + Bx* b  + Cx * c ]  / perimeter

Where Ax = the x coordinate  of A  Bx  = x coordinate of B    Cx  = x coordinate of C

And a, b , c  are  the sides lengths opposite A, B and C

So  we have

[ 5.5 (4)  + (4)(8)  + (0)(6)] / 18  =  3

Similarly the  y coordinate  of the center of  the incircle  =  

[Ay * a  + By * b + Cy * c ]  / 18 

[ 1.5sqrt (15)(4)  + (0)(8) + (0)(6)  / 18   =  sqrt (15)/3   =    radius  of incircle

Then the  height of triangle ANM = height of triangle ABC  - 2* incircle radius = sqrt (15)  ( 3/2 - 2/3)  =

(5/6) sqrt (15 )

And since triangles  AMN and ABC  are similar

Then

MN  / height of AMN  =  BC / height of ABC

MN  / [ (5/6)sqrt (15) ] =   4 /  [  (3/2 ) sqrt (15) ]

MN =  4 (5/6)  / (3/2)   =  4 ( 5/6) (2/3)  =  40 /18  =  20/9  

Hope this helps, whymenotsmart/(theisocelestriangle)^m^

Two right triangles have equal areas. The first triangle has a height of 5 cm and a corresponding base of 8 cm. The second triangle has a leg of length 20 cm. What is the length of the other leg of the second triangle?

5 * 8 = 20 * x

x = 2   

The length of the other leg is  2 cm

Proof:           5 * 8 / 2 = 20 cm²             2 * 20 / 2 = 20 cm²