Hello, orochimaru700!

Here's the first one . . .

1) Find the lateral area of a regular hexagonal pyramid

. . .if each edge of base is 10m and each lateral edge is 15m. I assume the "lateral edge" is what most of us call the "slant height".

One of the triangular side panels looks like this: Code:

*
/|\
/ | \
/ | \ 15
/ h| \
/ | \
/ | \
/ | \
* - - - * - - - *
5

From Pythagorus: .$\displaystyle h^2 + 5^2 \:=\:15^2 \quad\Rightarrow\quad h^2 \:=\:200 \quad\Rightarrow\quad h \:=\:10\sqrt{2}$

The area of this triangle is: .$\displaystyle \tfrac{1}{2}bh \:=\:\tfrac{1}{2}(10)\left(10\sqrt{2}\right) \:=\:50\sqrt{2}\text6{ m}^2$

Therefore, the area of the six panels is: .$\displaystyle A \;=\;6\times 50\sqrt{2} \;=\;300\sqrt{2}\text{ m}^2$