What is in-circle radius of the right triangle 3-4-5?

1) "A Pythagorean triple has three positive integers a, b, and c, such that a^2 + b^2 = c^2. In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c). Some well-known examples are (3, 4, 5) and (5, 12, 13)."
Source and further information:
http://en.wikipedia.org/wiki/Pythagorean_theorem


2) "We can draw a circle touching all 3 sides of any triangle, called the incircle with radius the inradius usually denoted by r and centre the incentre.
From the symmetry of the circle, a line from its centre to each vertex of the triangle will halve each of the angles in the triangle."

"Lines from the incentre to the vertices (shown in gray here) divide the triangle into three smaller ones, each having the same height, r on a base of one side of the whole triangle."

"The area of a triangle is one half of the base of the triangle times its height. So the three separate areas sum to the whole area:
area = a×r/2 + b×r/2 + c×r/2 = (a + b + c)×r/2

The sum of the sides of a triangle is the length of its perimeter and, since we have the perimeter halved in the formula we often find this expressed using the semiperimeter s:
area = inradius × semiperimeter = r×s "
Source and further information:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#incircle


3) For the right triangle 3-4-5, the area is the half product of the two legs (3×4/2=6), so we get:
area = inradius × semiperimeter
6 = inradius × (3+4+5)/2 = inradius × 6
This yields:
inradius = 1

in-circle radius of a right triangle ABC is its height (BD, not shown on the chart) to the hypothenuse AC:


A
|
|
|
| D
|___
B C

define BD as "a" and DC as "b", then
1) AD will be 5-b
2) a^2+b^2=3^2 and
3) a^2+(5-b)^2=4^2 => a^2+b^2-10b+25=16 => a^2+b^2-10b=-9
After combining 2) and 3) together, we get:
9-10b=-9 => b=1.8
Reassigning b=1.8 to 1), we get a^2+1.8^2=9 => a^2=5.76 => a=2.4
The answer: in-circle radius BD equals to 2.4