10.100. A regular hexagon $A B C D E F$ consists of two trapezoids sharing a common base $C F$ (Fig. 10.96). It is known that $A C=13$ cm, $A E=10$ cm. Find the area of the hexagon.
Solution.

$\triangle C K A$ is a right triangle, $A K=\frac{1}{2} A E=5$ (cm), $C K=\sqrt{A C^{2}-A K^{2}}=$ $=\sqrt{13^{2}-5^{2}}=12(\text{cm}), \cos \varphi=\frac{12}{13} ; \sin \varphi=\frac{5}{13}$. Consider $\triangle C O B(\angle C O B=$ $\left.=90^{\circ}\right): \sin 2 \varphi=\frac{5}{x}$ and $\sin 2 \varphi=2 \sin \varphi \cos \varphi=2 \cdot \frac{5}{13} \cdot \frac{12}{13}=\frac{120}{169}$. Therefore, $x=$ $=\frac{169 \cdot 5}{120}=\frac{169}{24}(\text{cm}), \quad \cos 2 \varphi=\frac{C O}{x} ; \cos 2 \varphi=\cos ^{2} \varphi-\sin ^{2} \varphi=\frac{144}{169}-$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-604.jpg?height=418&width=420&top_left_y=91&top_left_x=206)

Fig. 10.97

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-604.jpg?height=412&width=422&top_left_y=94&top_left_x=745)

Fig. 10.98

$-\frac{25}{169}=\frac{119}{169} \cdot$. Therefore, $C O=\frac{169}{24} \cdot \frac{119}{169}=\frac{119}{24}$ (cm), $C O=K F, C F=C O+O F=$ $=C K+K F=12+\frac{119}{24}=\frac{407}{24}$ (cm). The area of the desired hexagon: $S_{A B C D E F}=2 S_{C D E F}=2 \cdot \frac{1}{2}(C F+E D) \cdot E K=\left(\frac{407}{24}+\frac{169}{24}\right) \cdot 5=120\left(\text{cm}^{2}\right)$.

Answer: $120 \text{ cm}^{2}$.