# Math and Literature for Intermediate and Secondary

by Cindy Neuschwander

Hello, all!

I have attached a short list of some of my favourite math and literature connections for intermediate and secondary classes.  It follows on the heels of a workshop I gave yesterday in Maple Ridge, in which we explored important mathematical concepts in a series of engaging reads.  There is so much math potential in each of these stories that they can easily be shared with learners across the grades – either as a way to introduce a new topic or to present a context for a meaningful mathematical exploration.

I hope you find these titles  – and links to the mathematical concepts they address – helpful.

Enjoy!

Math and Literature for Intermediate and Secondary

# Explaining the derivation of the Area of a Circle – Grade 7

So… we all know that the formula of the area of a circle is  .

But have you ever considered why?  Students in Grade 7 need to understand the derivation of this formula and apply it to different situations.  My husband sent me these awesome on-line demonstrations of how the area of s circle is derived, using what we know about triangles.  Consider these images and what they show.  First, the circle is chopped up into roughly  triangular segments.

They are put together to form a parallelogram, in which the base is 1/2 of the circumference of the original circle.

The height of the parallelogram is the same as the radius of the circle (since each of the triangles is a section of the original circle).

To find the area of the parallelogram, we multiply the base times the height, or 1/2 of the circumference of the circle x the radius of the circle.  Consider it in this more abstract language:

1/2 ( ) r    which simplifies to

Cool, huh?

It’s WAY more effective to watch the video of the event. Check it out here!

Carole

PS – There’s another derivation that draws on the area of a triangle…  I won’t wreck it for you, but consider this… The base of the large triangle here is the whole circumference of the circle, or 2πr. Each of the little triangles that make it has a height the same as the radius of the original circle.  So that means that the area of this large circle is 1/2 base x height, or 1/2 (2πr) x radius/  Familiar?