In a parallelogram, diagonals 6√2 cm long and 14 cm long intersect at an angle of 450. Find the smaller side of the parallelogram.

1) by the property that the diagonals of the parallelogram are halved by the intersection point (O), we find halves of these diagonals (AO and BO), which with the smallest side of the parallel (AB) make up a triangle 14: 2 = 7 and 6√2: 2 = 3√2
2) by the cosine theorem, we find the smallest side. The AOB angle is 45 (cos45 = √2: 2) AB ^ 2 = AO ^ 2 + BO ^ 2-2 • AO • BO • cos