Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
Sorry I’m just commenting to do it
The answer is A because there are only acute angles
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So, therefore, C.
