7. $$AD = 2\sqrt{13}$$, $$DO = 4$$. Find the area of the side face, if the pyramid is regular?

Let $$ABCD$$ be a regular quadrangular pyramid with apex $$D$$ and base $$ABCD$$. Let $$O$$ be the center of the base. Then $$DO$$ is the height of the pyramid. Let $$AD = 2\sqrt{13}$$ and $$DO = 4$$. Let $$M$$ be the midpoint of the side $$BC$$. Then $$DM$$ is the apothem of the side face $$BCD$$. The area of the side face is $$S_{side} = \frac{1}{2} BC \cdot DM$$. Since $$O$$ is the center of the base, $$AO = OC = OB = OD$$. Triangle $$AOD$$ is right-angled. Using the Pythagorean theorem, we get $$AO = \sqrt{AD^2 - DO^2} = \sqrt{(2\sqrt{13})^2 - 4^2} = \sqrt{4 \cdot 13 - 16} = \sqrt{52 - 16} = \sqrt{36} = 6$$. Since $$AO = 6$$, then $$AC = 2 AO = 2 \cdot 6 = 12$$. Since the base is a square, $$AC = AB \sqrt{2}$$, then $$AB = \frac{AC}{\sqrt{2}} = \frac{12}{\sqrt{2}} = \frac{12 \sqrt{2}}{2} = 6\sqrt{2}$$. Thus, $$BC = 6\sqrt{2}$$. Since $$O$$ is the center of the base and $$M$$ is the midpoint of $$BC$$, $$OM = \frac{1}{2} AB = \frac{1}{2} \cdot 6\sqrt{2} = 3\sqrt{2}$$. Consider the triangle $$DOM$$. It is right-angled. Using the Pythagorean theorem, we get $$DM = \sqrt{DO^2 + OM^2} = \sqrt{4^2 + (3\sqrt{2})^2} = \sqrt{16 + 9 \cdot 2} = \sqrt{16 + 18} = \sqrt{34}$$. Now we can find the area of the side face $$BCD$$: $$S_{side} = \frac{1}{2} BC \cdot DM = \frac{1}{2} \cdot 6\sqrt{2} \cdot \sqrt{34} = 3 \sqrt{2 \cdot 34} = 3 \sqrt{68} = 3 \sqrt{4 \cdot 17} = 3 \cdot 2 \sqrt{17} = 6\sqrt{17}$$. Answer: $$6\sqrt{17}$$