The height of an isosceles trapezoid is 9 cm, and its diagonals are perpendicular. Find the perimeter of the trapezoid if its side is 12 cm.

AB = CD so the trapezoid is isosceles,
∠BAC = ∠CDA as angles at the base of an isosceles trapezoid,
AD – common side for triangles BAC and CDA, ⇒
ΔBAC = ΔCDA on both sides and the angle between them,
means ∠CAD = ∠BDA.
Then ΔAOD is isosceles rectangular.
ΔVOS is similar to it in two corners, which means it is also isosceles.
Let’s draw the height of the trapezoid KN through the point of intersection of the diagonals.
For isosceles triangles AOD and BOS, the OH and OK segments are heights and medians, and in a right-angled triangle, the median drawn to the hypotenuse is equal to its half:
KO = BC / 2
HO = AD / 2, ⇒
KH = (AD + BC) / 2 = 9 cm,
then AD + BC = 18 cm
Pabcd = 2AB + AD + BC = 24 + 18 = 42 cm