Area of a circle from what we already know

An interesting surprise

In this section we will explore how to determine the area of a circle by dividing it into sections and then rearranging these sections to form a shape with an easily recognizable area.[br]Follow the instructions and have fun!

1. Slider 1

The applet below shows a circle divided into 4 sections.[br]As we move [b]slider 1[/b] the number of sections increases by multiples of 4.[br]Move [b]slider 1[/b] to 3 and notice that there are now 12 sections in the circle.[br]We will explore what happens when we play with the other two sliders in the sections below.

2. Slider 2

Set [b]slider 1 [/b]to 5 (The circle will have 20 sections).[br][br]Move [b]slider 2 [/b]all the way to the left and describe what happens. [br]You can change the value of [b]slider 1[/b] to greater numbers to help in your description.

3. Slider 3

Now move [b]slider 3[/b] all the way down slowly and describe what happens once it gets to the bottom. Give as much detail as possible.

4. What shape do you see?

As [b]slider 3[/b] moved down with a large number of sections for [b]slider 1[/b], what shape is created by rearranging the sections?

A triangle A parallelogram Many triangles

5. Now, what is the area?

If you answered parallelogram in the previous question, you are correct![br]Write the formula for the area of a parallelogram.

6. Connecting to the circle part 1

The area of a parallelogram is [math]A=bh[/math] [br][br]The [i]height [/i]of the parallelogram [math]h[/math] is the same as...

the diameter of the circle the radius of the circle the circumference of the circle

7. Connecting to the circle part 2 - the base

The area of a parallelogram is [math]A=bh[/math][br][br]The applet shows the [i]base = [math]r\cdot\pi[/math][/i][br][br]Which of the following best describes why this is true?

The sections form a rectangle, not a parallelogram Because when the two sections slide to form the parallelogram, their total length is half of the circumference [math]C[/math], and since [math]C=2\pi r[/math], we have [math]\frac{2\pi r}{2}=\pi r[/math] Because the base of the parallelogram is the same as the circumference of the circle The applet is incorrect, the correct base should be [math]2\pi r[/math]

8. A little algebra

Simplify the right side of this equation. This is the area of the parallelogram:[br][br][math]A=r\cdot\pi\cdot r[/math][br][br]Comment on your simplified equation.