8.5
Circles
The problem of finding the area of a circle turns out to be very different from that of determining the area of a polygon. The ancient Egyptians had a formula (which we know about from Problem 50 of the Rhind Papyrus, dating from about 1650 BCE). They said that the area of a circle was the same as the area of a square whose side was eight-ninths of the circle's diameter. As we will see in the exercises, this is a very close approximation, but it is still an approximation.
One of the
standard problems in Greek geometry was
to construct, limited to the tools of straightedge and compass, a
square that had the same area as some particular given shape.
This is doable for regions whose edges are straight lines, but it is an
entirely different matter if the borders are curved. Hippocrates
of Chios (ca. 470 – 410 BCE) dealt with the shaded region in the figure
to the right (known as a lune). This is bordered by a semicircle
$\overparen{AEC}$and a quarter-circle $\overparen{AFC}$. Quite
remarkably, the area of this particular lune is the same as that of $\triangle ADC$, so it is possible to construct a square with the same area as this lune. Needless to say, Hippocrates' success in "squaring" this lune — with its borders that were arcs of circles — led others to try to construct squares with the same area as a whole circle. All such attempts failed, but it wasn't until 1882 that it was actually proved to be an impossible task. |
We could postulate the formula for the area of a circle, as we did the formula for the area of a square in the first section of this chapter. However, the following approach allows us to see where the formula comes from, though it does require a more intuitive approach than we have used previously.
We start with an important result about circles.
- Theorem 8.7: The ratio of the circumference to the diameter of any circle is a constant.
Proof: Consider two circles, one having radius $R$ and the other having radius $r$. Inscribe a regular $n$-gon in each circle (using the same $n$ in each case). The two $n$-gons are similar to each other, since the central angle for each of the smaller triangles in each one is $\left(\frac{360}{n}\right)^\circ$. We denote the perimeter of the polygon in the circle whose radius is $R$ by $P$ and the perimeter of the polygon in the circle with radius $r$ by $p$ (matching capital letters and lower case letters). Since the two $n$-gons are similar, we have $\frac{P}{R}=\frac{p}{r}$. Multiplying by $\frac{1}{2}$, we have $\frac{P}{2R}=\frac{p}{2R}$, or $\frac{P}{D}=\frac{p}{d}$. Now we increase the number of sides in the polygon, so that $n$ gets larger and larger. The ratios $\frac{P}{D}$ and $\frac{p}{d}$ change as $n$ increases, but since the polygons in the two circles are always similar to each other, the ratio $\frac{P}{D}$ always stays equal to $\frac{p}{d}$. As continues to increase, the perimeters of the two inscribed polygons approach the circumferences of the two circumscribing circles. It seems reasonable to conclude that $\frac{C}{D}=\frac{c}{d}$ for the two circles. That is, in any two circles, the ratio of the circumference to the diameter is a constant.
- Definition:
The constant $\dfrac{c}{d}$ is called $\pi$.
- Theorem 8.8:
The circumference $c$ of a circle of radius $r$ is $c=\pi d$ or $c=2\pi
r$.
We now turn our attention to the area of a circle.
- Theorem 8.9: The area $A$ of a circle of radius $\pi$ is $A=\pi r^2$.
When we replace the $c$ in this expression with $2\pi r$, we obtain $$\text{area(circle)}=\frac{1}{2}r c=\frac{1}{2}r(2\pi r)=\pi r^2.$$
Example: Determine the circumference and the areaof a circle of radius 5.
Solution: $c=2\pi r\quad \Rightarrow\quad c=10\pi$ And $a=\pi r^2\quad \Rightarrow \quad a=\pi\cdot 5^2=25\pi$