Question

A circle is drawn inside a right angle triangle whose sides are a, b, c where c is the hypotenuse, which touches all the sides of the triangle. Prove r = (a+b- c )/2 where r is the radius of the circle.

Answer 1

FIGURE IS IN THE ATTACHMENT

Let A, B and C be the sides of a right angled ∆ABC.
Let BC = a, AC = b , AB = c.
Let the circle touches the sides BC, CA, AB  at D, E and F.

Then, AE = AF & BD = BF & CD = CE
[Lengths of the tangent drawn from an exterior point to a circle are equal.]

In quadrilateral ABCD each angle is 90° , so OECD is a square.
OE = EC = CD = OD = r (radius)
AF = AE =(AC - EC ) = ( b -r )
BF = BD = (BC- DC) = (a - r)
AF + BF  = (b -r )+ (a - r)
AB = b - r + a - r    [AF + BF = AB]
c = a + b -2r           [AB = c]
2r = a + b - c

r = (a + b - c)/2

Hence, proved.

HOPE THIS WILL HELP YOU...

Answer 2

Solution :

-> Area of a Triangle in terms of inradius = rs ; [ where 's' = semi-perimeter ]
-> Area of this right angle triangle = 1/2 ( ab )

Equating both of 'em :
[tex]ab = r ( a + b + c ) \\ r = \frac{ab}{a+b+c} = \frac{ab(a+b-c)}{(a+b)^2 - c^2} = \frac{ab(a+b-c)}{(a^2 + b^2 - c^2 ) + 2ab} = \frac{a+b-c}{2}[/tex]

Note: We just use ( x + y )( x - y ) = ( x² - y² ) and that ( a² + b² - c² ) = 0