Hello, galactus!
Our areas agree, Soroban
. I included the center region.
I did, too, in my first run-through . . .
Hokceyman, the total length of all the tracks a bit more complicated.
Suppose the runner in the first lane runs along that inside edge.
His distance is the two straightaways and the two semicircles:
\(\displaystyle \;\;(2\,\times\,100)\,+\,2\pi\left(\frac{100}{\pi}\right) \:= \: 400\)
Suppose the runner in the second lane runs along
his inside edge.
His distance is the two straightaways (200 m) plus the slightly larger semicircles.
\(\displaystyle \;\;\)The radius is: \(\displaystyle \,\frac{100}{\pi}\,+\,1.\;\) The curved distance is: \(\displaystyle \,2\pi\left(\frac{100}{\pi}\,+\,1\right) \:=\:200\,+\,2\pi\)
His total distance is: \(\displaystyle \,200\,+\,(200\,+\,2\pi)\:=\:400\,+\,2\pi\)
Suppose the runner in the third lane runs along
his inside edge.
His distance is the two straightaways (200 ) plus the slightly larger semicircles.
\(\displaystyle \;\;\)The radius is: \(\displaystyle \,\frac{100}{\pi}\,+\,2.\;\) The curved distance is: \(\displaystyle \,2\pi\left(\frac{100}{\pi}\,+\,2\right) \:=\:200\,+\,4\pi\)
His total distance is: \(\displaystyle \,200\,+\,(200\,+\,4\pi)\:=\:400\,+\,4\pi\)
We find that each lane is \(\displaystyle 2\pi\) meters longer than the one before.
The ten lanes have lengths: \(\displaystyle \,400,\;400+2\pi,\;400+4\pi,\;400+6\pi,\;\cdots\;400+18\pi\)
This is an Arithmetic Series with first term \(\displaystyle a\,=\,400\), common difference \(\displaystyle d\,=\,2\pi\)
\(\displaystyle \;\;\)and \(\displaystyle n\,=\,10\) terms.
The sum of an Arithmetic Series is: \(\displaystyle \,S_n\;=\;\frac{n}{2}[2a\,+\,d(n-1)]\)
So we have: \(\displaystyle \:S_{10}\;=\;\frac{10}{2}[2\cdot400\,+\,(2\pi)(9) \;=\;4000\,+\,90\pi\) m.