The lengths of two adjacent sides of a parallelogram are 6 and 14. if the measure of an included angle is 60, find the length of the shorter diagonal of the parallelogram.

Let the name of the parallelogram be ABCD, the two adjacent sides of the parallelogram be AB and AD. AB=14 and AD= 6. The shorter diagonal of parallelogram is BD. The angle between AB and AD is 60.

In triangle ABD, according to the cosine rule,

BD^2=AB^2+AD^2-2(AB)*(AD) cos(60)

= 14^2+6^2-(2)14)(6)(0.5)

=196+36-84

= 232-84

=148

BD^2=148

BD=12.1655

So the length of the shorter diagonal is 12.1655 unit.

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