# Area of a circle sector

I have been given the following proportional relationship to derive the area of a circle's sector:

$\large\frac{\text{ Area of the sector}}{\text{Area of the circle}}=\frac{\text{arc length}}{\text{circumference of the circle}}$

From what I understand you're making equal the ratio of areas to the ratio of lengths. I know how to derive the area of a sector formula from this, that isn't my question. My question is why does this ratio hold in the first place?

If we were to describe the ratio of lengths and areas in general terms we would write: $\large \frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2$, where $A_1$ and $A_2$ are areas, and $l_1$ and $l_2$ are lengths. So with this in mind why isn't the $\frac{\text{s}}{\text{circumference of the circle}}$ side of the equation that consists of length, squared? Why isn't the expression my textbook gave: $\large\frac{\text{Area of the sector}}{\text{Area of the circle}}=\left(\frac{\text{arc length}}{\text{circumference of the circle}}\right)^2$ ?

• The ratio needs to be squared when you change the size in two dimensions. Here you keep the radius fixed.
– user65203
Commented Aug 8, 2014 at 17:10
• This is a bit jumping the gun, but is interesting to note that the ratio of the area to the total area (and so the arc length to the circumference) is the same as the ratio of the angle to the total radian measure of a circle. Commented Sep 23, 2018 at 6:48

Your ratios of lengths and areas description is true for plane figures that are similar to each other, in the sense of being the same shape but possibly different sizes.

Suppose a shape $S_1$ has an area $A_1$ and a length measured within/around it of $l_1$. Now suppose we dilate $S_1$ by a factor of $k$ to produce the new shape $S_2$ which is similar to $S_1$ and with area $A_2$ and matching length $l_2$.

Then $A_2 = k^2 A_1$ and $l_2 = k l_1$ and so it is true that $\frac{A_2}{A_1} = \left(\frac{l_2}{l_1}\right)^2$.

HOWEVER, circular sectors are NOT similar shapes. In the same circle, two circular sectors of different sizes are different shapes that are not similar. The relationship just mentioned therefore does not apply and in fact a different relationship applies.

This relationship actually comes from the relationship between the base and area of triangles of the same height...

Suppose you have two triangles $T_1$ and $T_2$ of the same height $h$ but different bases $b_1$ and $b_2$. Then the area of $T_1$ is $A_1 = \frac12 b_1 h$ and the area of $T_2$ is $A_2 = \frac12 b_2 h$. Therefore the ratio of the areas is: $$\frac{A_1}{A_2} = \frac{\frac12 b_1h}{\frac12 b_2 h} = \frac{b_1}{b_2}.$$ Because we are only changing one dimension and not the other, the ratio of the areas is the same as the ratio of the bases. With similar shapes, you change both dimensions simultaneously and so the area changes for both dimensions, hence the squaring.

Now consider what happens when you change the arc length of a circular sector: Imagine you have a sector and you change the angle by just a little bit. What you will add is a tiny sliver of area, which is very close to being a triangle. This triangle has base which is the tiny bit of extra arc length, and it has height equal to the radius. Each extra bit of arc length adds another triangular sliver of area, and all of these triangular slivers have the same height.

So in essence, extending the arclength of a circular sector is equivalent to extending the base of a triangle without changing its height. Therefore the ratio of areas will be the same as the ratio of arclengths. Basically by changing the arclength without changing the radius you are changing one dimension but not the other and so the ratio of areas will be only related to the ratio of the dimension you change.

• Thank you for your lovely answer. Commented Aug 17, 2014 at 13:17

It's not necessary to "correct for units" in this case - notice that $\frac{A_1}{A_2}$ divdes two areas. Looking at units, this is $\frac{\text{cm}^2}{\text{cm}^2}$, which means the answer is a number. So it has no units!

We have no problem talking about these two numbers being equal, no matter what units were used in the calculation - they disappeared when we divided on both sides.

Your intuition on the problem would be correct for a square, where we vary both side lengths at the same time. To follow from Yves Daoust's comment, this case is more similar to a rectangle where one side stays fixed (circumference of circle) and we vary the other side (s).

• I'm aware that when you divide two things with length you end up with a 'pure' number. What I'm confused with is why the ratio of areas is equal to the ratio of lengths? Shouldn't the ratio of the $\frac{A}{Area of the circle}$ be greater than $\frac{S}{Circumference of the circle}$ in this case? Commented Aug 9, 2014 at 11:14

You're used to the following relationship:

$$\frac{A_1}{A_2}=\left(\frac{\ell_1}{\ell_2}\right)^2$$

However, this relationship is typically only applied when we're talking about two separate similar shapes. For example, consider circles $-$ recall that any two circles are similar to each other. Thus, if we treat their circumferences as lengths and we let $r$ and $R$ denote the radii of the smaller and bigger circles respectively, then we obtain: $$\frac{\pi R^2}{\pi r^2}=\left(\frac{2\pi R}{2\pi r}\right)^2$$ which is indeed a true statement.

To see why the above relationship no longer holds for circle sectors, note that a circle sector is not similar to a circle. Indeed, both the circle sector and the entire circle have the same constant radius.

As a sanity check, consider a semicircle. Notice that its arc length is half that of the entire circumference. Indeed, its area is also half that of the entire circle (and not a quarter of a circle).

However, suppose that we didn't cut the circle down the middle in half. Instead, consider what would happen if we scaled the entire circle by a factor of a half so that its radius is divided by two. Then this new circle would indeed be a quarter of the original circle.

• I was reading over questions I've asked previously, and your answer now comes across as extremely lucid, thank you for your help. Commented Jan 25, 2015 at 16:51

You're dealing with ratios of areas and ratios of lengths, rather than areas and lengths.

The relation $A \propto l^2$ that you observe is indeed true, but when expressing the area ratio this $l^2$ dependence is in both the numerator and the denominator. So, there's no need to take account of it again.