please help!

I'm assuming that you want the radius of the inscribed circle.

The only way that I can see to find the radius of the inscribed circle is to use coordinate geometry and a very nasty formula; so, if anyone else has a nicer solution, please post it!

A 3-4-5 triangle is a right triangle and the center of the inscribed circle is the point of intersection of the angle bisectors.

Let A = (0,3), B = (4,0), and C = (0,0).

The angle bisector of angle(C) is the line whose equation is y = x.

To find the angle bisector of angle(B), we can use this formula (where m = slope of the bisector,

                m₁ = slope of one of the sides, m₂ = slope of the other side):

      m =  [ m₁·m₂ + sqrt( 1 + m₁² )·sqrt(1 + m₂² ) - 1 ] / [ m₁ + m₂ ]

The slope of BC = 0; the slope of AB = (0 - 3) / (4 - 0) = -3/4

     m =  [ (0)·(-3/4) + sqrt( 1 + (0)² )·sqrt(1 + (-3/4)² ) - 1 ] / [ (0) + (-3/4) ]

         =  [ 0 + sqrt( 1 )·sqrt( 25/16 ) - 1 ] / [ -3/4 ]

         =  -1/3             <---  this is the slope of the angle bisector at vertex B

The equation of the angle bisector at B is:  y - 0  =  (-1/3)(x - 4)     --->     y  =  (-1/3)x + 4/3

The two lines:  y = x   and   y  =  (-1/3)x + 4/3   intersect at th point  (1,1) 

because:                              x  =  (-1/3)x + 4/3

                                     (4/3)x  =  4/3

                                             x  =  1

The distance from the point (1,1) to the x-axis is 1; as is the distance to the y-axis and also to line AB.

Thus, the radius of the incircle is 1.

A triangle has side lengths 3, 4, and 5.  Find the radius of the inscribed (triangle) circle.

Semiperimeter is:         s = (3+4+5)/2 = 6

Area of a triangle is:     A = 1/2 (3*4) = 6 u²

Inradius r is:                     r = A / s = 1