I am pleased to say that — beyond spending every day on the water this summer — I DID manage to create a new teacher resource for my intermediate colleagues.
This time, it’s a stand up calendar of problems — one for every day of the school year!
This compact but potent book comes with an easel so you can set it up on your desk and flip from one rich problem to the next, posing open-ended questions of your intermediate students.
Good Questions: A Year of Open-Ended Tasks is a problem-a-day resource that includes
rich tasks ideal for grades 5, 6, 7 and 8. Organized by topic and structured in problem sets of 5 or more, this simple to use teacher resource includes 210 mathematically important questions to engage your students in deep thinking. For only $25, it’s a perfect back-to-school gift for yourself!
Proportional reasoning, measurement, operations and algebra are featured in this calendar of problems. Each one engages students in thinking flexibly, critically and creatively in the face of important and challenging mathematics.
Let the problem-solving begin!
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
It’s WAY more effective to watch the video of the event. Check it out here!
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?