**Q) **If a, b and c are the sides of a right angled triangle, where c is hypotenuse, then prove that the radius of the circle which

touches the sides of the triangle is given by r =

**Ans: **

Let’s consider a right angled triangle ABC with sides a, b & c.

Its ∠ A is right angle and hence, side c is hypotenuse.

We draw the circle of radius R inside the circle.

It touches side AB at point D, side AC at point E and side BC at point F.

Next, we connect Point O with Points D, E and F

Since the Length of the tangents drawn from an external point are equal

∴ AD = AE, BD = BF, and CF = CE

In Quadrilateral ADOE,

AD = AE (tangents from same external point)

OD = OE (radii of same circle)

∠ODA = ∠OEA = 90

∴ ADOE is an square with sides = r

∴ Side a = AE + CE = r + CE

∴ CE = a – r ………. (ii)

Similarly, Side b = AD + BD = r + BD

∴ BD = b – r ………. (iii)

Now side BC = BF + CF

Since BF = BD and CF = CE (tangents from same external point)

∴ BC = BD + CE

substituting values from equations (ii) and (iii):

∴ c = (b – r) + (a – r)

∴ c = a + b – 2 r

∴ 2 r = a + b – c**∴ r = **

**Hence Proved!**