A Circle Inscribed in a Triangle

$OP \perp AB$ $OM \perp AC$ $ON \perp CB$ Diagram labelled. Point P is a the intersection of the hypotenuse and a line drawn from the circle center to it. $MOCN$ is a square of side $R$ . Inspecting the diagram, one sees that $\triangle MOA$ and $\triangle OPA$ are congruent. Therefore:

$AM = AP = 3-R$

Again, $\triangle ONB$ and $\triangle OBP$ are congruent. Therefore:

$NB=PB=4-R$

Therefore, the hypotenuse is:

$AB = AP+PB = 3-R + 4-R = 5$ $\implies \boxed{R=1}$

I think you are enjoying doing animation! Area of the triangle is $\triangle =\dfrac{1}{2}\times 3\times 4=6$ ,

semi-perimeter is $s=\dfrac {3+4+5}{2}=6$ .

So radius of the incircle is $R=\dfrac{\triangle }{s}=\boxed 1$ .

If a circle is inscribed in a triangle, the radius of the circle is given by

$R = \dfrac{2A}{P}$ where: $\text{A=area of the triangle}$ and $\text{P= perimeter of the triangle}$

The area of the triangle is $A = 0.5\times 3\times 4 = 6$ . The perimeter is $P=3+4+5=12$ . So the radius of the inscribed circle is

$R = \dfrac{2\times 6}{12}=\dfrac{12}{12} = \large{\boxed{1}}$ $\boxed{\text{answer}}$

Smallest pythagorean triplet, thus smallest integer incircle radius.