An isosceles trapezoid with an acute angle of 60 is described near a circle with a root radius of 3.

An isosceles trapezoid with an acute angle of 60 is described near a circle with a root radius of 3. Find the midline of the trapezoid.

From point O, the center of the circle, draw the radii OK and OM to the swing points of the circle and the sides of the CD and AD.

Since the points K and M are points of tangency, and the segments DK and DМ are tangent, then the segment OD is the bisector of the angle ADC, then the angle MDO = ADC / 2 = 60/2 = 30.

In a right-angled triangle ОDM, the leg ОМ = 3 cm as the radius of the circle and lies against the angle 30, then ОD = 2 * ОМ = 2 * √3 cm.

Then, by the Pythagorean theorem, DM ^ 2 = OD ^ 2 – OM ^ 2 = 12 – 3 = 9.

DМ = 3 cm.

In a right-angled triangle ABН, the leg BН = 2 * √3 as the diameter of the inscribed circle.

Then AH = BH / tg60 = 2 * √3 / √3 = 2 cm.

AM = DM = 3 cm, then НM = AM – AH = 3 – 2 = 1 cm.

DН = DМ + НМ = 3 + 1 = 4 cm.

In an isosceles trapezoid, the BH height divides the larger base into two segments, the larger of which is equal to the length of the midline.

Answer: The length of the middle line of the trapezoid is 4 cm.