Why pi?

[rhu]

My son asked a great question the other day: Why is pi between 3 and 4?

It's easy to show why 4 is an upper bound on pi, by inscribing a unit circle in a unit square.

But I have not yet been able to come up with an explanation of why 3 should be a lower bound for pi. Inscribing a square inside the unit circle gives 2.8+, and I suppose I could try higher-order polygons, but does anyone out there have a demonstration that will resonate with a fifth-grader?

Comments

[thedan.livejournal.com]

Inscribe a hexagon into a unit circle, and split the hexagon into six equilateral triangles of sidelength 1.

If you start at the leftmost point and follow the circle around to the rightmost point, the length of the arc is pi (since the circle has circumference 2 pi). But if you follow the edges of the triangles, which is clearly a shortcut, the total length is 3.

[thedan.livejournal.com]

Though to be fair, this argument relies on the assumption that you can fit equilateral triangles together to make a hexagon. But if you accept that the angles of a triangle add up to 180 degrees, you can probably get there intuitively without trig.

[steveanastasi.livejournal.com]

Great question. I made this learning packet to show why pi has to be more than three but less than four. Enjoy!

Why is pi between three and four? (http://www.sophia.org/packets/why-is-pi-between-three-and-four)

http://www.sophia.org/packets/why-is-pi-between-three-and-four (http://www.sophia.org/packets/why-is-pi-between-three-and-four)

[cnoocy]

Constructing the hexagon with a compass, while not mathematically rigorous, does illustrate that the hexagon is composed of 6 segments of length 1.

[rikchik.livejournal.com]

Inscribing a hexagon should work. Radius of 1, circumference of 6.

[drdaly.livejournal.com]

The area of an inscribed hexagon is 3 (three rhombuses with sides of length 1), your desired integer lower bound.