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## 'Of All the Areas' printed from http://nrich.maths.org/

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:
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.