10.041. The measure of one of the angles of a 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 $\sqrt{75} / 2 \mathrm{~cm}$. Find the lengths of the sides and the longer diagonal of the parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-575.jpg?height=420&width=648&top_left_y=92&top_left_x=106)

Fig. 10.41

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-575.jpg?height=302&width=572&top_left_y=193&top_left_x=664)

Fig. 10.42
## Solution.

Draw $B N \perp A D$ (Fig. 10.41); since $B N=2 O M$, then $B N=\sqrt{75}$ cm. Considering that in $\triangle A N B \quad \angle A B N=30^{\circ}$, 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. Next, from $\triangle B N D$ we get $N D^{2}=B D^{2}-B N^{2}=124-75=49$; 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: $12 ; 10$ and $2 \sqrt{91}$ cm.