In parallelogram ABCD AB = 13 cm, AD = 37 cm, angle B = 130 degrees. Find the diagonal BD.

Diagonal BD divides the parallelogram into two equal triangles – ABD and BCD. Consider a triangle ABD. Two sides are known in it: AB = 13 cm and AD = 37 cm. The sum of the angles in the parallelogram adjacent to one side is 180º. Find angle A:

180º – 130º = 50º.

By the cosine theorem, the side of BD equals

BD = √ (AB² + AD² – 2 * AB * AD * cosA).

Let’s find this side:

BD = √ (13² + 37² – 2 * 13 * 37 * cos 50º) = √ (169 + 1369 – 962 * cos 50º) ≈ √ (1538 – 680.24) ≈ √857.76 ≈ 29.29 cm.

Answer: The diagonal BD of parallelogram ABCD is approximately 29.29 cm.