4.19. Find the area of the total surface of a cone if its lateral surface can be unfolded into a sector with a radius of 1 and a right central angle.

## Group 5
4.19. The sector of the cone's development has an angle of $90^{\circ}$ and radius $R=1$; therefore, its area $S_{1}=\pi R \cdot \frac{1}{4}=\frac{\pi}{4}$. The lateral surface area of the cone $S_{2}=\pi r l=\pi r$ (since $l=R=1$). But $S_{1}=S_{2}$, so $\frac{\pi}{4}=\pi r$, from which $r=\frac{1}{4}$. Now we find the total surface area of the cone:

$$
S=S_{2}+S_{\text {base }}=\frac{\pi}{4}+\frac{\pi}{16}=\frac{5 \pi}{16}
$$

Answer: $\frac{5 \pi}{16}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-581.jpg?height=334&width=600&top_left_y=952&top_left_x=46)

Fig. 4.17