A certain right-angled triangle has an area $t=121.5 \mathrm{~cm}^{2}$; one of its angles $\alpha=36^{\circ} 52^{\prime} 10.7^{\prime \prime}$; calculate the surface area and volume of the double cone that is formed by the triangle's rotation around its hypotenuse. $\pi=3.14159$.
$$
\begin{aligned}
& F=2 t \pi(\sin \alpha+\cos \alpha)=2 t \pi \sqrt{2} \sin \left(45^{\circ}+\alpha\right)=1068.75 \text{ cm}^{2} \\
& V=\frac{2 \pi t}{3} \sqrt{t \sin 2 \alpha}=2748.2 \text{ cm}^{3}
\end{aligned}
$$