Question

If a,b,c are the sides of right angled triangle, where c is hypotenuse then prove that the radius r of the circle which touches the sides of a triangle is given by :-

r = a + b - c / 2

Answer 1

the circle touches the sides  AB,BC and CA of triangle ABC at D, E and F
AD=AF,  BD=BE  and CE=CF
Similarly EB=BD=r
Then we have  c = AF+FC
                   ⇒ c = AD+CE
                   ⇒ c = (AB-DB)(CB-EB)
                   ⇒ c = a-r +b-r
                   ⇒ 2r =a+b-c
                        r =( a+b-c)/2

Answer 2

Answer:

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

Step-by-step explanation:

Given: a,b,c are sides of right triangle & circle of radius r touches the sides of triangle.

To prove: $r=\frac{a+b-c}{2}$.

Let us first draw the figure:

Assume circle touches the sides of circle $AB,BC,CA$ at $P,Q,R$ respectively.

Also we know that length of tangents from external point are equal.

$AR=AP\\BQ=BP\\CR=CQ=r$

By figure,

$AP=b-r\\BP=a-r$

$AB=AP+BP\\c=b-r+a-r\\c=a+b-2r\\2r=a+b-c\\r=\frac{a+b-c}{2}$

Hence proved that radius of circle is $r=\frac{a+b-c}{2}$.

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