In the given figure, ABC is a right triangle right-angled at

From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, we have

BQ = BP

Let us denote BP and BQ by x

AP = AR

Let us denote AP and AR by y

RC = QC

Let us denote RC and RQ by z

We have been given that is a right triangle and BC = 6 cm and AB = 8 cm. let us find out AC using Pythagoras theorem. We have,

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2}=6^{2}+8^{2}$

$A C^{2}=36+64$

$A C^{2}=100$

$A C=\sqrt{100}$

$A C=10$

Consider the perimeter of the given triangle. We have,

AB + BC + AC = 8 + 6 + 10

AB + BC + AC = 24

Looking at the figure, we can rewrite it as,

AP + PB + BQ + QC + AR + RC = 24

Let us replace the sides with the respective x, y and z which we have decided to use.

$y+x+x+z+y+z=24$

$2 x+2 y+2 z=24$

$2(x+y+z)=24$

$x+y+z=12$.....................(1)

Now, consider the side AC of the triangle.

AC = 10

Looking at the figure we can say,

AR + RC = 10

y + z = 10 …… (2)

Now let us subtract equation (2) from equation (1). We have,

x + y + z = 12

y + z = 10

After subtracting we get,

x = 2

That is,

BQ = 2, and

BP = 2

Now consider the quadrilateral BPOQ. We have,

BP = BQ (since length of two tangents drawn to a circle from the same external point are equal)

Also,

PO = OQ (radii of the same circle)

It is given that .

From the property of tangents, we know that the tangent will be at right angle to the radius of the circle at the point of contact. Therefore,

$\angle O P B=90^{\circ}$

$\angle O Q B=90^{\circ}$

We know that sum of all angles of a quadrilateral will be equal to . Therefore,

$\angle P B Q+\angle O P B+\angle O Q B+\angle P O Q=360^{\circ}$

$90^{\circ}+90^{\circ}+90^{\circ}+\angle P O Q=360^{\circ}$

$270^{\circ}+\angle P O Q=360^{\circ}$

$\angle P O Q=90^{\circ}$

Since all the angles of the quadrilateral are equal to and the adjacent sides also equal, this quadrilateral is a square. Therefore, all sides will be equal. We have found out that,

BP = 2 cm

Therefore, the radii

PO = 2 cm

Thus the radius of the incircle of the triangle is 2 cm.