Question

In right angled triangle ABC, angle ACB =90°. A circle is inscribed in the triangle with radius r.a,b,c are the lengths of the sides BC, AC and AB respectively. Prove that 2r=a+b-c

Answer 1

Answer:hope it helps buddy! if so please mark it as brainliest_/\_

Step-by-step explanation:

since these points produce tangents at the circle, hence

AP = AQ

CP = CR

BR = BQ

Given,

AB = c

AC = b

BC = a

since CPOR is a square hence,

CP = CR = r

=> BR = BC - CR = a - r

=> BQ = BR = a - r

similarly,

AP = AC - CP = b - r

=> AQ = AP =  b - r

Now,

AB = c

=> AQ + BQ = c

=> b - r + a - r = c

=> b + a - 2r = c

=> 2r = a + b - c

which is the required equation.!!!!!!!!!!!

Answer 2

Answer:

We have proved that 2r=a+b-c for a figure in which a circle is inscribed inside a triangle.

Step-by-step explanation:

Due to the fact that these locations cause tangents to the circle,

AP = AQ

CP = CR

BR = BQ

Given,

AB = c

AC = b

BC = a

Since CPOR is a square hence,

CP = CR = r

So, BR = BC - CR = a - r

=> BQ = BR = a - r

Similarly,

=> AQ = AP =  b - r

Now,

AB = c

=> AQ + BQ = c

Thus,  c =  b - r + a - r  

This implies c = b + a - 2r

Hence,  a + b - c = 2r

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