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In the given figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.

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#### Solution

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 _{Δ}ABC is a right triangle and *BC* = 6 cm and *AB* = 8 cm. let us find out *AC* using Pythagoras theorem. We have,

`AC^2=AB^2+BC^2`

`AC^2=6^2+8^2`

`AC^2=36+64`

`AC^2=100`

`AC= sqrt100`

`AC=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`

`2x+2y+2z=24`

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

`x+y+z=12`

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 `∠PBQ= 90^o`

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,

`∠OPB= 90^o`

`∠OQB= 90^o`

We know that sum of all angles of a quadrilateral will be equal to `360^o`. Therefore,

`∠PBQ+∠OPB+∠OQB+∠POQ=360^o`

`90^o + 90^o +90^o + ∠POQ= 360^o`

`270^o + ∠POQ = 360^o`

`∠ POQ= 90^o`

Since all the angles of the quadrilateral are equal to `90^o`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.

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