Sorry to be so long in replying, but it took me a while to "see" this the right way....
I'm sure you've drawn the square-based right pyramid. Let's label the height as "h", the slant height (that is, the height of a triangular face) as "s", and the length of a side of the base as "b".
Looking at a cross-sectional view of the pyramid, so you're looking at an isosceles triangle with base b and height h, the height-line splits this triangle into two right triangles, each having a base with length b/2 and a height of h. The Pythagorean Theorem gives you the length of the hypotenuse:. . . . .
If you look at the pyramid from another angle, you'll note that the above also stands for the slant-height s, being the distance from the peak of the pyramid, right down the middle of a side, to the midpoint of that side of the base.
Then the area of a triangular face is given by:. . . . .
You are given that this area (above) is equal to the square of the height h, so:. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .
But of course the square must be positive, so you can ignore the "minus" on the square root, giving you:. . . . .
Leaving that aside for now, let's turn to the ratio, using the radical expression found previously for the slant-height:. . . . .. . . . .
The square root in the numerator can be replaced, because we saw above that:. . . . .
So we get:. . . . .
Now replace b2
with the solution value (from the quadratic) to get:. . . . .. . . . .
The ratio is now strictly numerical, which is promising, but you need to "rationalize" to get rid of the radicals in the denominator, so:. . . . .
Simplify, and see what you get.