The sum of the acute angles of the trapezoid is 90 degrees, the height is 2, and the base is 4 and 8.

The sum of the acute angles of the trapezoid is 90 degrees, the height is 2, and the base is 4 and 8. Help to find the sides of the trapezoid.

ABCD – trapezoid, AD = 8, BC = 4, angle A + angle D = 90 degrees, BH = 2 – height.
1. The height BH divides the base AD into two segments, one of which is equal to the half-sum of the bases, and the second is the half-difference of the bases. Then:
AH = (AD – BC) / 2 = (8 – 4) / 2 = 4/2 = 2.
2. Consider the triangle AHB: angle AHB = 90 degrees (since BH is the height), BH = 2 and AH = 2 are legs, AB is the hypotenuse. Since BH = AH = 2, then AHB is an isosceles triangle: BH and AH are the lateral sides, AB is the base, the angles HAB and HBA are the angles at the base:
HAB angle + НВA angle + AНВ angle = 180 degrees;
x + x + 90 = 180;
2x = 90;
x = 45.
HAB angle = HBA angle = 45 degrees.
Angle HAB = angle A.
By the Pythagorean theorem:
AB = √ (AH ^ 2 + BH ^ 2) = √ (2 ^ 2 + 2 ^ 2) = √ (4 + 4) = √8 = 2√2.
3. Since angle A + angle D = 90 degrees, then:
45 + angle D = 90;
angle D = 45 degrees.
4. From vertex C we draw the height of the CK to the base AD: KD = 2, CK = 2, angle D = 45 degrees, then the triangle CKD is isosceles. СK = DK = 2.
By the Pythagorean theorem:
CD = √ (CK ^ 2 + DK ^ 2) = √ (2 ^ 2 + 2 ^ 2) = √ (4 + 4) = √8 = 2√2.
Answer: AB = 2√2, CD = 2√2.