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=2x+y−z.
Get the complete, step-by-step math solution for: "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 ...". Powered by SolveForX AI math tutor.
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=2x+y−z.