The diagonals of the trapezoid are mutually perpendicular and equal to 8 cm and 6 cm.

The diagonals of the trapezoid are mutually perpendicular and equal to 8 cm and 6 cm. Find the middle line of the trapezoid.

Let a trapezoid ABCD be given, the diagonals of which AC and BD are mutually perpendicular and AC = 8 cm, BD = 6 cm.

From vertex C we draw a straight line CK parallel to BD, we get a quadrilateral BCKD, which is a parallelogram, since:

* CK || ВD (by construction),

* ВC || DK (the segments lie on the parallel bases of the trapezoid ABCD).

In the parallelogram BCKD, the sides are CK = BD = 6 cm and DK = BC.

In triangle ACK, side AK = AD + DK = AD + BC. The angle ACK is straight, since CK || BD and AC ⊥ BD, which means AC ⊥ CK.

By the Pythagorean theorem AK ^ ² = AC ^ ² + CK ^ ², we obtain:

AK ^ ² = 8 ^ ² + 6 ^ ²;

AK = 10 cm or AD + BC = 10 cm.

The middle line of the trapezoid MN is equal to the half-sum of the bases of the trapezoid:

MN = (AD + BC): 2; MN = 10: 2 = 5 (cm).

Answer: The middle line of the trapezoid is 5 cm long.