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Jane noticed the patterns in the not tilted triangles:

They are square numbers, the little one is area $1$, then area $4$, then $9$ going up in size accross the given image.
This is surprising as they are areas of triangles, and they are giving square numbers. You can calculate the size of a triangle by counting the length of a side and squaring it.

Esther deduced the following after looking at lots of examples of tilted triangles.

The formula for the area of a tilted triangle is based around $n^2$.
If the amount of tilt is $x$, and the place in the sequence is $n$, the area is $n^2 +xn +x^2$ triangular units.

Adele from Coventry High School used the image below to prove Esther's conjecture:

Generalised tilted triangle
We have that $x$ is the tilt of the small triangle. So the edge length of the big triangle is $2x+n$, so the area is $(2x+n)^2$ from Jane above.
We then want to find the area of the bits between the big and little triangles.
So if we look just in the bottom right of the picture as the three sections are the same.
We have an equilateral triangle in the corner with edge length $x$, so that area is $x^2$.
For the area of the other triangle, we cannot use standard formula for area as we have unit tiangles rather than unit squares. If we look at that area on the isometric paper we get a parallelogram with edges $n$ and $x$, which means the parallelogram has area $2nx$ because of the isometric nature. The area of that scalene triangle is then $nx$, so the whole thing is $nx+x^2$.
The area of the little triangle is
$A=(2x+n)^2-3\times (nx+x^2)$
$A=4x^2 + 4nx + n^2 - 3nx - 3x^2$
$A=x^2 + nx + x^2$
So Esther was right.