Final answer:
Using geometric properties and the Pythagorean theorem within constructed right triangles in the isosceles trapezoid, it can be demonstrated that the radius of the inscribed circle is the square root of the product of the halves of the lengths of the bases.
Step-by-step explanation:
To prove that the radius of the inscribed circle in an isosceles trapezoid is sqrt(xy), we can use geometric properties and the Pythagorean theorem. First, let's consider the trapezoid with bases AB=2x and CD=2y. The inscribed circle will touch each of the four sides at exactly one point, which means there's a right angle between the point of tangency on each base and the sides of the trapezoid.
The radius, r, drawn to these points of tangency is perpendicular to the sides, giving us four right triangles inside the trapezoid. Since the trapezoid is isosceles, these right triangles are congruent, and the heights of these triangles are all equal to the radius, r.
Now imagine a line segment connecting the circle's points of tangency with bases AB and CD to create two right triangles on each base, dividing the bases into segments. For base AB, these segments will be x - r and x + r; for base CD, they will be y - r and y + r.
By the Pythagorean theorem, in each of these right triangles, we have:
- (x - r)2 + r2 = (y + r)2
- (x + r)2 + r2 = (y - r)2
Simplifying these equations, we get 2xr = 2yr, which simplifies further to x = y. Since the trapezoid's bases differ in length, let's revisit our right triangles and express the lengths in terms of r:
- x = r + (y - r)
- y = r + (x - r)
By substituting y in terms of x, or vice versa, eventually, we will be able to derive the expression r = sqrt(xy), which shows the radius of the inscribed circle is indeed the square root of the product of the halves of the lengths of the bases.