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'Triangles All Around' printed from http://nrich.maths.org/
Sam sent us her work on the problem,
including the angles of all the triangles. Thank you
Can you see how she avoided counting any
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
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
For the eight point board, I again labelled the pegs
There are five possible triangles
- ABC, with angles $135^\circ$, $22.5^\circ$ and
- ABD, with angles $112.5^\circ$, $22.5^\circ$ and
- ABE, with angles $90^\circ$, $22.5^\circ$ and
- ACE, with angles $90^\circ$, $45^\circ$ and $45^\circ$.
- ACF, with angles $45^\circ$, $67.5^\circ$ 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