1.58. Perpendiculars are drawn from the vertex of the acute angle of a rhombus to the lines containing the sides of the rhombus to which this vertex does not belong. The length of each perpendicular is 3 cm, and the distance between their bases is $3 \sqrt{3}$ cm. Calculate the lengths of the diagonals of the rhombus.
1.58. Since $\triangle A E F$ is isosceles (Fig. 1.45), the bisector $A M$ is perpendicular to $E F$ and lies on the diagonal of the rhombus. We find

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-439.jpg?height=523&width=498&top_left_y=1336&top_left_x=55)

Fig. 1.44

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-439.jpg?height=479&width=525&top_left_y=1368&top_left_x=626)

Fig. 1.45
$A M^{2}=A F^{2}-M F^{2}=9-\frac{27}{4}=\frac{9}{4}$, i.e., $A M=\frac{3}{2}$ (cm). In $\triangle A C F$, we have $\angle F=90^{\circ}$ and $F M \perp A C$; therefore, $A F^{2}=A C \cdot A M$, or $9=A C \cdot \frac{3}{2}$, from which $A C=6$ (cm). Further, $\triangle A C D \sim \triangle A E F$ (the angles at the base are equal as angles with mutually perpendicular sides) and, thus, $\frac{A M}{O D}=\frac{E F}{A C}$, or $\frac{\frac{3}{2}}{O D}=\frac{3 \sqrt{3}}{6}$, from which $O D=\sqrt{3}$ cm. Therefore, $B D=2 \sqrt{3}$ cm, $A C=6$ cm.

Answer: $2 \sqrt{3}$ and 6 cm.