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
,
and
.
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
,
and
.
- ABD, with angles
,
and
.
- ACE, with all angles
(an equilateral triangle).
For the eight point board, I again labelled the pegs ABCDEFGH
There are five possible triangles
- ABC, with angles
,
and
.
- ABD, with angles
,
and
.
- ABE, with angles
,
and
.
- ACE, with angles
,
and
.
- ACF, with angles
,
and
.
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