10.036. A rhombus with a side length of 6 cm is inscribed in a right-angled triangle with a $60^{\circ}$ angle, such that the $60^{\circ}$ angle is common to both and all vertices of the rhombus lie on the sides of the triangle. Find the sides of the triangle.
Solution.

In $\triangle F B D$ (Fig. 10.36) the leg $F D=6$ cm and lies opposite the angle of $30^{\circ}$, from which $F B=12$ cm and, therefore, $A B=18$ cm. Next, in $\triangle A B C$ we have $A C=A B / 2=9$ cm. Finally, $B C=\sqrt{A B^{2}-A C^{2}}=\sqrt{243}=9 \sqrt{3}$ cm.

Answer: $9, 9 \sqrt{3}$, and 18 cm.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-573.jpg?height=384&width=384&top_left_y=78&top_left_x=176)

Fig. 10.37

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-573.jpg?height=298&width=530&top_left_y=163&top_left_x=687)

Fig. 10.38