1.28. One of the angles of the parallelogram is $60^{\circ}$, and the shorter diagonal is $2 \sqrt{31}$ cm. The length of the perpendicular dropped from the point of intersection of the diagonals to the longer side is $\frac{\sqrt{75}}{2}$ cm. Find the lengths of the sides and the longer diagonal of the parallelogram.
### 1.28. Let's conduct

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-426.jpg?height=548&width=413&top_left_y=111&top_left_x=732)

Fig. 1.23 $B N \perp A D$ (Fig. 1.24); since $B N=2 O M$, then $B N=\sqrt{75}$ cm. Considering that in $\triangle A N B \angle A B N=30^{\circ}$, we have $A B=2 A N$ and, therefore, $4 A N^{2}=75+A N^{2}$, from which $A N=5$ cm and $A B=10$ cm. Further, from $\triangle B D N$ we get

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-426.jpg?height=359&width=633&top_left_y=732&top_left_x=518)

Fig. 1.24 $N D^{2}=B D^{2}-B N^{2}=124-75=49$; hence, $N D=7$ cm and $A D=12$ cm. Finally, from the equality $A C^{2}+B D^{2}=2 A B^{2}+2 A D^{2}$ we find $A C^{2}=200+288-124=364$, i.e., $A C=2 \sqrt{91}$ cm.

Answer: 10 and 12 cm; $2 \sqrt{91}$ cm.