Teaching maths so it counts in real life

- Group B

As we have seen how we could prove that the sum of the interior angles of a regular polygon would sum up to (n-2) x 180 degrees based on the information that the sum of exterior angles always sums up to 360 degrees.

We thought it would be nice to share what an alternative proof (hopefully it holds for all cases). This alternative involves cutting any n-sided polygons into triangles with a vertex of the polygon arbitrarily chosen as a common vertex for all the triangles to be formed. Examples of 4-sided, 6-sided and 8-sided polygons are shown in Figure 1 below.

Figure 1

As we observe from these “cut” polygons, it is obvious that the resulting number of triangles formed is always 2 less than the number of vertices initially present, i.e., for a n-sided polygon, there will be (n-2) triangles being formed.

Tapping on the prior knowledge that the sum of the interior angles of a triangle is 180 degrees, the sum of interior angles of a n-sided polygon would thus be (n-2) x 180.

Group A

Pictures on Algetiles

Dear all,

From trying to brainstorm for a consistent approach using algetiles...

To our brilliant hypotheses!

Group D

Group A

Group C

Group E? F?

Group B?

Blurred shot...

Group A

And our ever photo-ready Liting. =)

It was fun! Looking forward to more! =)

God Bless!
Regards,
Daniel