12.127. The development of the lateral surface of a cylinder is a rectangle, the diagonals of which intersect at an angle $\alpha$. The length of the diagonal is $d$. Find the lateral surface area of the cylinder.
## Solution.

Let $A A_{1} B_{1} B$ be the development of the lateral surface of the cylinder, $A A_{1}$ and $B B_{1}$ be its generators, $A B$ and $A_{1} B_{1}$ be the lengths of the circumferences of its bases, $A B_{1}=d, \angle A O A_{1}=\alpha$ (Fig. 12.127). Since $\angle A O A_{1}$ is the external angle of the isosceles triangle $A O B$, we have $\angle O B A=\frac{\alpha}{2}$. From $\triangle A_{1} A B$ we find $A_{1} A=d \sin \frac{\alpha}{2}, A B=d \cos \frac{\alpha}{2}$. We obtain

$$
S_{6 \mathrm{ox}}=2 \pi R H=A B \cdot A A_{1}=d^{2} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}=\frac{1}{2} d^{2} \sin \alpha
$$

Answer: $\quad \frac{1}{2} d^{2} \sin \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-812.jpg?height=320&width=474&top_left_y=192&top_left_x=155)

Fig. 12.127

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-812.jpg?height=464&width=438&top_left_y=32&top_left_x=729)

Fig. 12.128