# Given: A parallelogram with diagonals equal to 56cm and 34cm. The intersect of the angles is 120^circ. Find the shorter side's length.

Question
Given:
A parallelogram with diagonals equal to 56cm and 34cm. The intersect of the angles is $$\displaystyle{120}^{\circ}$$. Find the shorter side's length.

2021-03-09
Suggest the length of the shorter side is x. The $$\displaystyle{120}^{\circ}$$ angle is opposite to the longer side. So the angle opposite to the shorter side is $$\displaystyle{180}-{120}={60}^{\circ}$$. Now we have a triangle with one side being x, and the two others: 17cm and 28cm, because the diagonals of a parallelogram bisect each other. Use the Law of Cosines to solve this.
$$\displaystyle{x}^{{2}}={28}^{{2}}+{17}^{{2}}-{2}{\left({28}\right)}{\left({17}\right)}{{\cos{{60}}}^{\circ}}$$
$$\displaystyle{x}^{{2}}={597}$$
$$\displaystyle{x}=\sqrt{{597}}$$
$$\displaystyle{x}\approx{24.4}$$ cm
The shorter side is approximately 24.4cm long.

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