Hello, IDontunderstand!

If you don't know Trig, we can still work it out.

The base of the pyramid is an equilateral triangle with length of 6

The height of the pyramid is 9

How do I find the slant height of the pyramid?

The base is an equilateral triangle of side 6.

Code:

A
*
/|\
/ | \
/ | \
6 / |h \ 6
/ | \
/ | \
/ | \
B * - - - + - - - * C
3 D 3

In right triangle $\displaystyle ADC\!:\;\;h^2 +3^2 \:=\:6^2 \quad\Rightarrow\quad h \:=\:3\sqrt{3}$

We want the "center" of the triangle (the centroid $\displaystyle G$).

It divides an altitude in the ratio 2:1

Code:

A
*
/|\
/ | \
/ | \
/ | \
/ | _ \
/ |2√3 \
/ | \
/ G* \
/ | _ \
/ |√3 \
/ | \
B * - - - - - - - - - - - * C

This is the point directly below the top vertex of the pyramid $\displaystyle (V)$.

Now consider the side view of the pyramid.

Code:

V
*
|\
| \
| \
9 | \ s
| \
| \
| \
G * - -_- * A
2√3

We have: .$\displaystyle s^2 \:=\:9^2 + (2\sqrt{3})^2 \:=\:81 + 12 \:=\:93$

Therefore, the slant height is: .$\displaystyle s \:=\:\sqrt{93}$