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 = \frac{a + b - c}{2}

If a, b and c are the sides of a right angled triangle
CBSE Sample paper 2024


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 = \frac{(a + b - c)}{2}

Hence Proved!

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