In an isosceles trapezoid, the lengths of the bases are 6 cm and 4 cm, and the length of the height is 4 cm.

In an isosceles trapezoid, the lengths of the bases are 6 cm and 4 cm, and the length of the height is 4 cm. Find the radius of the circle described around this trapezoid. It is known that the center of the circle lies inside the trapezoid.

From points B and C, draw parallel perpendiculars ВН and СK (heights), then the quadrangle НВСК is a square, since ВН = СK = 4 cm and ВС = НК = 4 cm.
The segments AH and KD will be equal:
АН = KD = (АD – НК) / 2 = (6 – 4) / 2 = 2/2 = 1 (cm).
Triangles AHB and DKC are equal. Consider the triangle AHB: HH = 4 cm – height, therefore the angle AHB = 90 degrees, therefore AH = 1 cm and BH = 4 cm – legs, AB – hypotenuse. By the Pythagorean theorem:
AB = √ (AH ^ 2 + BH ^ 2)
AB = √ (1 ^ 2 + 4 ^ 2) = √ (1 + 16) = √17 (cm).
The radius of the circle described by the isosceles of this trapezoid is calculated by the formula:
R = c√ ((ab + c ^ 2) / (4c ^ 2 – (a – b) ^ 2)),
where a is a larger base, b is a smaller base, c is a side.
The radius of a circle circumscribed about a trapezoid, given by condition, is:
R = √17 * √ ((6 * 4 + (√17) ^ 2) / (4 * (√17) ^ 2 – (6 – 4) ^ 2)) = √17 * √ ((24 + 17) / (4 * 17 – 2 ^ 2)) = √17 * √ (41 / (68 – 4)) = √17 * √ (41/64) = √17 * √41 / √64 = √17 * √41 / 8 = √697 / 8 (cm).
Answer: R = √697 / 8 cm.