11.012. The diagonal of the square base of a regular quadrilateral pyramid is equal to its lateral edge and is equal to $a$. Find the total surface area of the pyramid and its volume.
## Solution.

Let the side of the square be denoted by $b$. The diagonal of the square $d = b \sqrt{2}$. It is equal to the lateral edge of the pyramid $a = b \sqrt{2}$ (Fig. 11.10). The area of the base $S_{\text{base}} = b^2 = \frac{a^2}{2}$. The height of the pyramid, as seen from the diagram, is $h = \sqrt{a^2 - \left(\frac{d}{2}\right)^2} = \sqrt{a^2 - \left(\frac{a}{2}\right)^2} = \frac{a \sqrt{3}}{2}$.

The slant height of the pyramid $l = \sqrt{h^2 + \left(\frac{b}{2}\right)^2} = \sqrt{\frac{3a^2}{4} + \frac{a^2}{8}} = \frac{a \sqrt{7}}{2 \sqrt{2}}$. The semi-perimeter of the base $p = 2b = \frac{2a}{\sqrt{2}}$. The lateral surface area $S_{\text{lateral}} = l \cdot p = \frac{a \sqrt{7}}{2 \sqrt{2}} \cdot \frac{2a}{\sqrt{2}} = \frac{a^2 \sqrt{7}}{2}$. 

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-657.jpg?height=560&width=562&top_left_y=96&top_left_x=139)

Fig. 11.11

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-657.jpg?height=556&width=438&top_left_y=102&top_left_x=765)

Fig. 11.12

The total surface area $S = S_{\text{lateral}} + S_{\text{base}} = \frac{a^2 \sqrt{7}}{2} + \frac{a^2}{2} = \frac{a^2 (\sqrt{7} + 1)}{2}$.

The volume $V = \frac{1}{3} S_{\text{base}} \cdot h = \frac{1}{3} \frac{a^2}{2} \cdot \frac{a \sqrt{3}}{2} = \frac{a^3}{4 \sqrt{3}} = \frac{\sqrt{3} a^3}{12}$.

Answer: $\frac{a^2 (1 + \sqrt{7})}{2} ; \frac{\sqrt{3} a^3}{12}$.