Sam sent us her work on the problem,
including the angles of all the triangles. Thank you
Sam!

Can you see how she avoided counting any twice?

She first counted triangles with two corners on neighbouring pegs, then those two apart, and so on. She called two triangles the same if they had the same angles and not just if they used the same pegs. She explains this in a bit more detail below.

Here is her work:

First I labelled the points on the pegboard $ABCD$.

There are four possible triangles: $ABC$, $ABD$, $ACD$ and $BCD$. However these triangles are all the same shape (you can see this by rotating triangle $ABC$) so we could say that there is only one type of triangle that we can make.

This triangle has angles $90^\circ$, $45^\circ$ and $45^\circ$. I know this because if you draw a square around the points $ABCD$ and cut it in along the diagonal you get this triangle.

For the six point board, I again labelled the points as $ABCDEF$.

There are three possible triangles

Can you see how she avoided counting any twice?

She first counted triangles with two corners on neighbouring pegs, then those two apart, and so on. She called two triangles the same if they had the same angles and not just if they used the same pegs. She explains this in a bit more detail below.

Here is her work:

First I labelled the points on the pegboard $ABCD$.

There are four possible triangles: $ABC$, $ABD$, $ACD$ and $BCD$. However these triangles are all the same shape (you can see this by rotating triangle $ABC$) so we could say that there is only one type of triangle that we can make.

This triangle has angles $90^\circ$, $45^\circ$ and $45^\circ$. I know this because if you draw a square around the points $ABCD$ and cut it in along the diagonal you get this triangle.

For the six point board, I again labelled the points as $ABCDEF$.

There are three possible triangles

- $ABC$, with angles $120^\circ$, $30^\circ$ and $30^\circ$.
- $ABD$, with angles $90^\circ$, $60^\circ$ and $30^\circ$.
- $ACE$, with all angles $60^\circ$ (an equilateral triangle).

For the eight point board, I again labelled the pegs
$ABCDEFGH$

There are five possible triangles

- ABC, with angles $135^\circ$, $22.5^\circ$ and $22.5^\circ$.
- ABD, with angles $112.5^\circ$, $22.5^\circ$ and $45^\circ$.
- ABE, with angles $90^\circ$, $22.5^\circ$ and $67.5^\circ$.
- ACE, with angles $90^\circ$, $45^\circ$ and $45^\circ$.
- ACF, with angles $45^\circ$, $67.5^\circ$ and $67.5^\circ$.

Can you see how she worked out the angles in the triangles? If you have come across circle theorems you may find these helpful. Remember that the angles in a triangle add up to 180 degrees! You can divide the triangle (or the circle) into pieces whose angles you know to help you.

If you would like to have a go at this problem for yourself, you might like to print off these sheets if you're not using the interactivity:

Sheet of four-peg boards

Sheet of six-peg boards

Sheet of eight-peg boards