Final answer:
The proof for the radius 'r' of an inscribed circle within a right-angle triangle with sides 'a', 'b', and hypotenuse 'c' leverages the concept of the triangle's area expressed in two ways - through its sides and through the semiperimeter and the radius of the inscribed circle. By equating these two, we can deduce the formula r = (a+b-c)/2.
Step-by-step explanation:
The student has asked to prove that for a circle inscribed in a right-angle triangle with sides 'a', 'b' and hypotenuse 'c', the radius 'r' of the circle is given by the formula r = (a+b-c)/2. To begin with, we acknowledge the Pythagorean theorem, which states that in a right-angle triangle, the sum of the squares on the two shorter sides is equal to the square on the hypotenuse (a² + b² = c²). However, this theorem is not directly used to prove the formula for 'r'.
Instead, we use the fact that the area of the triangle can also be expressed using the radius of the inscribed circle. The area 'A' of the right-angle triangle is equal to (1/2)ab, but it can also be expressed through the triangle's semiperimeter 's' (which is (a+b+c)/2) and the radius 'r' of the inscribed circle as A = r(s). By setting these two expressions for the area equal, we have (1/2)ab = r((a+b+c)/2). Simplifying this, we can solve for the radius 'r', which yields r = (a+b-c)/2.
This relation shows that the radius of the inscribed circle is directly linked to the sides of the right-angle triangle in a very elegant mathematical relationship.