About sum of right-angled triangles'area and area of circle.

Question

the image shows right-angled triangles in semi-circle

In Definite Integration, we know that area can be found by adding up the total area of each small divided parts.

So, base on the Definite Integration, we may say the area of circle is equal to $\sum^{n}_{k=1}(h_k)$

And we also know that $A_k=2r(h_k)(1/2)=rh_k$ while $A$ is refer to the right-angled triangle's area.

so that, $A_k(1/r)=h_k$

As a result this equation comes out with fixed position of diameter:

$\sum^{n}_{k=1}(h_k)=\pi r^2$

$\sum^{n}_{k=1}(A_k)(1/r)=\pi r^2$

$\sum^{n}_{k=1}(A_k)=\pi r^3$

I wish to know whether I am correct or not, thanks.

Answer 1

You have wrongly assumed that the sum of the heights $h_k$ is equal to the area of the circle. Area has dimensions $[L^2]$, so the area of the circle is actually the sum of the areas of the rectangular strips of height $h_k$ and infinitesimal base length $dx$.

$h_k=A_k/r\\\displaystyle\sum_1^\infty h_kdx=\sum_1^\infty\frac{A_kdx}r=\pi r^2$

Answer 2

No, you are all wrong.

If you want to make a sum of areas of some triangles that is approximate to the area of the circle, these triangles must not overlap. On your linked picture the two triangles obviously overlap (have some common area), so the further discussion is pointless.