# Drawing a regular pyramid

I've been asked for help in high school mathematics (some basic stereometry) but I'm not sure how to solve this exercise:

Draw a regular triangular pyramid given the lengths of edges $3.8\ cm$ and $4.5\ cm$ (first length is the length of side of base).

I assume one should construct an isosceles triangle but how does one find the fourth vertex? I think any point works, but am not sure about this. (I vaguely remember we might have done this in high school, but can't remember exactly how - I've done only pure mathematics since.)

Personally, I'd try to draw this in perspective, but I'm in serious doubt that this is how they do it at that level.

Now, I've taken courses in synthetic projective and non-Euclidean geometry, but not in any descriptive geometry, and I'm basically clueless how these problems are solved. If you could point me to some good material, I would be very appreciative.

EDIT: I know how to calculate all the lengths by using Pythagoras theorem. I want to know how to use these measurements to draw a proper picture. To be more precise:

1) If I construct isosceles triangle using given measurements, how do I find the fourth point? I believe it depends on perspective, but let's ignore it for the moment (this is a high school question). Is the fourth point drawn by some convention, if we ignore perspective? Is this a correct approach at all?

2) If we do not ignore perspective, how do I calculate vanishing points or in broader sense, what do I need to be aware of? You don't need to explain in detail, a reference will be fine. (note: I have master's degree in pure mathematics, I can probably read any source you can throw at me)

$$l^2 + (a/2)^2 = L^2 ,l^2 - (a/2)^2 = h^2$$
where $a = 3.8 , L = 4.8 =$ corner to vertex of pyramid and $l$ is slant triangle mid-altitude. These basic dimensions help further in connecting to the vanishing point after a choice of eye height.