In the given figure, x, y and z are the sides of a right triangle, where z is the hypotenuse. Prove that the radius r of the circle which touches the sides of the triangle is given by $r = \frac{x + y - z}{2}$ .

Answer: The radius $r$ of the inscribed circle is $r = \frac{x + y - z}{2}$ .

Step-by-step solution

Step 1: Identify Tangent Properties

Let the right triangle be $ABC$ , with the right angle at $C$ . Let the sides opposite to vertices $A$ , $B$ , and $C$ be $x$ , $y$ , and $z$ respectively. The circle is inscribed in the triangle, touching sides $BC$ , $AC$ , and $AB$ at points $P$ , $Q$ , and $R$ respectively. Since tangents from an external point to a circle are equal in length, we can establish relationships between the segments. $CP$ and $CQ$ are radii to the points of tangency from the right angle vertex, forming a square $CQOP$ (where $O$ is the center of the circle). Thus, $CQ = CP = r$ . The length $BP$ is $BC - CP = x - r$ . Similarly, $AQ$ is $AC - CQ = y - r$ . Also, $BR = BP = x - r$ and $AR = AQ = y - r$ .

Step 2: Relate Tangent Segments to Hypotenuse

The hypotenuse $AB$ is composed of the segments $AR$ and $BR$ . We can express $AB$ in terms of $x$ , $y$ , and $r$ using the tangent segment lengths identified in the previous step.

Step 3: Substitute and Simplify

Substitute the expressions for $AR$ and $BR$ into the equation for the hypotenuse $AB$ . This gives $z = (y - r) + (x - r)$ . Simplify the equation to $z = x + y - 2r$ .

Step 4: Solve for Radius r

Rearrange the equation to solve for $r$ . First, move $2r$ to one side and $z$ to the other: $2r = x + y - z$ . Then, divide by 2 to isolate $r$ , which yields the desired formula $r = \frac{x + y - z}{2}$ .

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In the given figure, x, y and z are the sides of a right triangle, where z is the hypotenuse. Prove that the radius r of the circle which touches the sides of the triangle is given by r=x+y−z2r = \frac{x + y - z}{2}r=2x+y−z​.