## Task A-3.2.

The base of a regular right quadrilateral pyramid is a square with a side length of 12, and the height is 8. What is the surface area of this pyramid?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-13.jpg?height=528&width=811&top_left_y=2055&top_left_x=545)

If $a$ is the side of the base, $v$ is the height of the pyramid, and $h$ is the height of the slant face of the pyramid, then according to the Pythagorean theorem,

$$
\left(\frac{a}{2}\right)^{2}+v^{2}=h^{2}
$$

Thus, $h=\sqrt{6^{2}+8^{2}}=10$.

The surface area of the pyramid is $4 \cdot \frac{12 \cdot 10}{2}+12^{2}=384$.