Triangles

In right triangle ABC, AB=10, AC=6 and BC=8 units. What is the distance from C to the midpoint of segment AB ?

Here's another way

The area of the triangle  is   (1/2) product of the legs  = (1/2) (6) (8)  = 24 units^2

So....calling AB the base....the altitude, h,  is : 

24  = (1/2) (10) h

24/5  =  h

Call the altitude, CD

So.....

AD  = sqrt [ AC^2  - CD^2] =

sqrt (6^2  - (24/5)^2)   =

sqrt  (36  - 576/25 ]  =

sqrt [ 324/25]  =

18/5

So....DM  = 5 - 18/5  =  7/5

So  CM  = 

sqrt  ( DM^2  + CD^2 ] = 

sqrt [ (7/5)^2  + (24/ 5)^2 ]  = 

sqrt [ 49  + 576] / 5 =

sqrt [625] /5  =

25 / 5   =

5