Join our community!

Update all PDFs

Wedges of a Circle


Alignments to Content Standards: 7.G.B.4

Task

Martin and Muriel finished a project for class showing one way to see why the area of a circle is given by $A=\pi r^2$, if $r$ is the radius of the circle. Muriel is not in class today and Martin is trying to understand the following page of pictures from their project.



Help Martin by writing up an explanation of how these pictures could be used to derive the formula for the area of a circle.  

IM Commentary

The purpose of this task is to help students visualize one way to derive the formula for the area of a circle, which uses the formulas for the circumference of a circle and the area of a parallelogram. The standard asks for "informal" derivations of this formula, and this task lets students both play and convince themselves. Instructors are encouraged to help any struggling students continue the process by making more drawings (with increasing number of wedges) or by more carefully labeling pieces of the drawing (e.g., the radius). Alternatively, a more hands-on version of this task can be made by having students cut circles into wedges and reassemble the wedges themselves. An outline can be found on the NCTM illuminations page.

Solution

We see that as we decompose the circle into a larger number of smaller wedges, the reassembled shape begins to more closely resemble that of a parallelogram. If we consider continuing this process, then the reassembled wedges will even more closely resemble a "complete" parallelogram. As we have not removed or added any area in our process, the areas of the circle and the parallelogram should be equal. Thus, we derive the area of the circle by instead finding the area of the parallelogram.

Let $r$ be the radius of the circle and $C$ its circumference. Then the area of the parallelogram is given by $A=\frac{1}{2}C r$, since the height of the parallelogram is $r$ and the width is comprised of half of the circumference of the circle. Since $C=2\pi r$, we have 

$$A=\frac{1}{2}Cr = \frac{1}{2}(2\pi r) r = \pi r^2.$$