Advanced 2D Geometry Websites for Deeper Understanding

[z-component]

Does anyone have any useful websites for 2D geometry that discuss some more advanced concepts for a person interested in more of the "backbone" of geometry?

For example, instead of what the usual high school geometry course covers, some more in-depth knowledge of geometry would be great!

 

[Werg22]

Concerning geometry I have tought of something. Per exemple to calculate the volume of a "perfect" pyramid, you would have to imagine an infinite number of square composing it. The area of a square decreases with the height of the triangle. It is easy to find a function of the area of the square depending on this lenght. Graphing this function would give you a curve, and intergrating the area from the area point (0,0) to (final lenghts, final area) would perhaps give you the right result. The reason I think it would be correct is that if you calculate the area, you would have to find a function of the perimeter instead of the area. Since this function is linear, I was able to find that it was correct...

 

[amcavoy]

Concerning geometry I have tought of something. Per exemple to calculate the volume of a "perfect" pyramid, you would have to imagine an infinite number of square composing it. The area of a square decreases with the height of the triangle. It is easy to find a function of the area of the square depending on this lenght. Graphing this function would give you a curve, and intergrating the area from the area point (0,0) to (final lenghts, final area) would perhaps give you the right result. The reason I think it would be correct is that if you calculate the area, you would have to find a function of the perimeter instead of the area. Since this function is linear, I was able to find that it was correct...

I don't think that is the question though. I mean, you could find the volume of a perfect pyramid with something like the following:

\sum (h-y)\Delta y=\int_{0}^{h}(h-y)^{2}dy=\frac{h^{3}}{3}

...but without calculus, how would it be done? Is that what you're asking?