The base of pyramid is regular hexagon with side a=4 cm. Each lateral face is isosceles triangle with base a=4cm. You have to find the height of this triangle in order to count the whole lateral area.
Combine the center of hexagon with hexagon's vertex and consider the right triangle formed by half of side base (first leg), half of hexagon's height (second leg) and segment that goes from the center of hexagon to the vertex (hypotenuse). By the Pythagorean theorem,
[tex] 2^2+h^2=4^2,\\ h^2=16-4,\\ h^2=12,\\ h=2\sqrt{3} [/tex] cm.
Now consider right triangle formed by height h=[tex] 2\sqrt{3} [/tex] cm, height of the pyramid H=8 cm and height l of lateral face. By the Pythagorean theorem,
[tex] (2\sqrt{3}) ^2+8^2=l^2,\\ l^2=12+64,\\ l^2=76,\\ l=2\sqrt{19} [/tex] cm.
Now you can find the area of one lateral face:
[tex] A_{l}=\dfrac{1}{2} \cdot a\cdot l=\dfrac{1}{2} \cdot 4\cdot 2\sqrt{19}=4\sqrt{19}[/tex] sq. cm.
The whole lateral area of the pyramid is:
[tex] A_{lateral}=6\cdot A_l=6\cdot 4\sqrt{19}=24\sqrt{19}=104.6 [/tex] sq. cm.
Answer: 104.6 sq. cm.
