Find the midline of an isosceles trapezoid whose diagonals are perpendicular and its height is 1dm.

From the vertex C of the trapezoid, draw a segment CE parallel to the ВD.

Since, by condition, the diagonals intersect at an angle of 90, the ACE angle of the ACE triangle is also equal to 90.

In an isosceles trapezoid, the diagonals at the intersection point are divided into equal segments, ОА = ОD, then the triangle AOD is isosceles, and the angle OAD = 45.

Then the ACE triangle is rectangular and isosceles, CA = CE. The height CH, drawn from the vertex of the right angle ACE, divides the hypotenuse AE into equal segments, AH = EH. Then, by the property of the height of a right-angled triangle, CH^2 = AH * EH = AH^2.

AH^2 = 12.

AH = 1 dm, then AE = 2 * 1 = 2 dm.

AE = AD + DE.

DE = BC, since ВСЕD is a parallelogram, and its opposite sides are equal.

AE = AD + BC, which is the sum of the grounds.

Then the middle line is:

MP = (AD + BC) / 2 = AE / 2 = 2/2 = 1 dm.

Answer: The middle line of the trapezoid is 1 dm.