When i last analyzed the properties of similar triangles, i thought i was toiling a bit too much over a simple fact... But i now see that was time well spent, for the properties of similar triangles seem to be a master stroke of geometry without which many pre-calculas era derivations wouldn't be possible.
And, for that matter, even in the calculus era, one would not be able to tell between linear and non-linear functions without using the properties of similar triangles. (i'll let you think through why, in case it isn't obvious)
Here, lets see how wonderfully the properties of similar triangles become an essential ingredient in the derivation for the cone volume.
We won't go all the way till the cone volume, but pause at an essential step in getting there.
Before getting to cone, lets look at a pyramid, it looks some thing like below with a triangular base area of 'b', and 3 triangular sides meeting at an apex at height 'h':
We can view this pyramid as a stack of infinitesimally thin triangles laid one on top of the other with continuously decreasing size till they converge at the apex.
Now, if the height of the pyramid is fixed, is there any relationship between two triangles?: i) the ABC at the base (b), and ii) the other, A'B'C' at some arbitrary height h' with area (b') ?
The first observation is that triangles ABC and A'B'C' are "similar" (similar triangle usage count: 1)
How? There are many ways to get to it, but here is a simple one:
a) ABC and A'B'C' are parallel to each other across the z-axis, which means all corresponding sides of the two triangles are parallel to each other. (this is how we laid out pyramid, by placing one triangle on top of the other)
b) Lets (just pick up and) drop lines C'A' and B'C' onto the x-y plane. C'A' is parallel to CA and B'C' is parallel to BC. So, the angle we get when we make lines CA and BC intersect is same as the angle we get when we make the lines C'A' and B'C' intersect.
c) So ∠BCA = ∠B'C'A', and on similar lines we can establish that the other two corresponding angles are equal.
As stated earlier, we assumed that area of triangle ABC = b and that of A'B'C' = b'.
One property of similar triangles is that the ratio of their areas is equal to the ratios of the squares of any of their corresponding sides.
Why is that so?
Lets quickly examine this from the below figure where triangles ABC and EBD are similar to each other:
Area(ABC)/Area(EBD) = (b*h/b`*h`) [The halves in the formula cancel out]
But, we also see that traingles AGC and EFD are similar too, and hence h/h` = c/c`.
(similar triangle usage count: 2)
But since ABC and EBD are similar, c/c` = b/b`, and hence:
Area(ABC)/Area(EBD) = (b*h/b`*h`) = (b/b`)*(h/h`) = (b/b`)*(c/c`) = (b/b`)*(b/b`) = (b*b)/(b`*b`)
Now lets us go back to our first figure:
So we have it that Area(ABC)/Area(A`B`C`) = b/b` = (AB*AB)/(A`B`*A`B`).
A cool trick or two from here gets us to note this beautiful fact that the ratio can be completely expressed in terms of the triangles's corresponding height from the x-y plane.
trick #1) Observe that triangles ABV and A`B`V are similar too.. This means (AB/A`B`) = (AV/A`V)
(similar triangle usage count: 3)
Next lets drop three lines on our figure:
i) a line parallel to z-axis through point A
ii) a line parallel to z-axis through point A`, and
iii) a line parallel to x-axis through point V
We get a figure as below:
trick #2: Observe that triangles AVW and A`VW` are similar. So, AV/A`V = AW/A`W` = h/(h-h`)
(similar triangle usage count: 4)
So, Area(ABC)/Area(A`B`C`) = (AB*AB)/(A`B`*A`B`) = (AV*AV)/(A`V*A`V) = (h*h)/( (h-h`)*(h-h`) )
Area(ABC)/Area(A`B`C`) = (h*h)/( (h-h`)*(h-h`) )
Now, here lets pause at this remarkable fact we understood: if we have a pyramid with a base area b and height h, the area of every triangle in the pyramid parallel to the base, along the z-axis shrinks exclusively as a function of how high up the triangle is.
This leads to another remarkable fact, but we need to consider the volume of pyramid first.
The volume of the pyramid is the sum of all the stacked up triangles.
So, lets assume we divided the pyramid ABCV into some n infinitesimally small triangles.
Consider another pyramid XYZT with same height h and same base area b but a completely dissimilar base shape.
Now, this new pyramid can also be divided into the same n infinitesimally small triangles.
If we start at the base, base_area(ABC) = base_area(XYZ) = b
Then when we go to the first decrement in ABCV, A`B`C`:
Area(A`B`C`) = [Area(ABC) * (h-h`)*(h-h`) ] / (h` * h`) = [b * (h-h`)*(h-h`) ] / (h`*h`)
Now, lets get to the corresponding first decrement in XYZT, X`Y`Z`:
Area(X`Y`Z`) = [Area(XYZ) * (h-h`)*(h-h`) ] / (h` * h`) = [b * (h-h`)*(h-h`) ] / (h`*h`)
Which means even if we have two dissimilar pyramids but with same base area b and height h, their volumes will be equal.
Isn't it an amazing property to just stand in awe of ? and why not thank the properties of similiar triangles, which we used a total 5 times in this thought process.
