This is a homework question I am struggling with...

Let $n$ be a positive integer, and cut the circle into $n$ equal sectors. In each sector there is an isosceles triangle formed where the edges of the sector intersect the circle. Prove the formula for the area of the circle is correct by taking the sum of the areas of the triangles without bound.

I've gotten as far as figuring out the central angle of each triangle is $\dfrac{2 \pi}{n}$. I'm having trouble with getting the area of the triangle into a form that I can take the integral of.

  • 2
    $\begingroup$ You seem intereted in applying differential calculus. Instead you should be thinking about applying integral calculus: taking a limit of the sum of a lot of tiny areas is what Riemann sums and Riemann integrals are all about. $\endgroup$ – Lee Mosher Apr 29 '15 at 14:46
  • $\begingroup$ woops! not sure why I wrote derivative there. I meant integrals. I should have waited until I finished my coffee to write that up. $\endgroup$ – abstractbryan Apr 29 '15 at 14:50
  • 1
    $\begingroup$ What you need is limits, not derivatives. Figure out the area of your triangle. You will get an expression containing the sine of some small angle. Multiply that expression by $n$, since there are $n$ such triangles. Then take the limit $n \rightarrow \infty$ of the resulting expression (sum), keeping in mind that $\sin(\theta) \rightarrow \theta$ as $\theta \rightarrow 0$. $\endgroup$ – inwit Apr 29 '15 at 15:02
  • $\begingroup$ math.stackexchange.com/questions/720935/… $\endgroup$ – user117644 Mar 2 '16 at 3:14

Let $r$ be the radius of the circle. The area of the triangle is $\frac12 (base \times height)$, if you split each of the triangles down the middle (to give a right angled triangle) then the height will be given by $$height = r \cos(\tfrac\pi n),$$ notice how we have halfed the angle as we split the triangle in 2. Similarly the base will be given by $$ base = 2 r \sin(\tfrac\pi n), $$ where we have remembered to multiply by two as the base of the right angled triangle is only half that of the isoceles.

Putting these quantities into the formula for the area of the isoceles triangle we get \begin{align*} A_{triangle} &= \frac 12 \times 2 r \sin(\tfrac\pi n) \times r \cos(\tfrac\pi n) \\ &= r^2 \sin(\tfrac\pi n)\cos(\tfrac\pi n) \end{align*}

Now there are n triangles so the total area of all the triangles would be given by $$ A_{total} = r^2 n\sin(\tfrac\pi n)\cos(\tfrac\pi n) , $$ which can be rewritten using the double angle formula as $$ A_{total} = r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}). $$ The area of the circle can then be found if we take the limit as $n\rightarrow\infty$. $$ A_{circ} = \lim \limits_{n \rightarrow \infty} r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}), $$ and using the fact that for very small $x$, $\sin(x) \approx x$ this can be seen to be \begin{align*} A_{circ} &= \lim \limits_{n \rightarrow \infty} r^2\frac{2\pi n} {2n} \\ &=\pi r^2, \end{align*} which is precisely what we want.

  • $\begingroup$ You used a lot of facts that you didn't even point out you were using, such as: the perpendicular you constructed does indeed divide the triangle into two parts of equal area; the perpendicular bisects the base of the triangle; the perpendicular does indeed bisect the central angle. Also an explicit mention of the use of $\lim_{x \to 0} \frac{\sin x}{x}$ would have been useful. $\endgroup$ – Junglemath Oct 4 '19 at 1:53

In addition to Rammus' answer there are less trigonometric and more concise ways of solving the problem. I will show below three methodologies: with limits, with a modification of Riemann Sums (notice that the question asks for a sumation), and finally how to express the same modified Riemann Sum as an integral.

First to the intuition. You can approximate the area of the circle by dividing it into isosceles triangles.

Source: http://www.amsi.org.au/teacher_modules/the_circle.html

Notice the gap between the base of the triangles and the arc of the circle. That is the error in our approximation. By increasing the number of triangles, that region decreases. So if we increase our (n) number of triangles to infinity, the error disappears, and the base of a given triangle is equal to the arclength of a given segment.

(1) Note the equation for the area of an isosceles triangle and the equation for circumference of a circle:

$$A_{triangle} = \frac 12 \times base \times height \\ C = 2 \pi r$$

When $n = \infty$:

$$height = r, \space base = \frac {2\pi r} n$$

Since we have $n$ triangles the total area can be expressed like this:

$$A_{circ} = \lim \limits_{n \rightarrow \infty} n \space (A_{triangle}) = \lim \limits_{n \rightarrow \infty} n \space \frac 12 \frac {2\pi r} n r$$

The n's cancel and so do the 2's, giving us: $A_{circ} = \pi r^2$

(2) Though, let's take this a step farther and see if we can rearrange this into something resembling a Reimann Sum, which works by summing infinitely many rectangles under a given function $f(x)$. It's form is:

$$ A = \lim \limits_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_{i-1}) \Delta x \\ \Delta x = \frac {b-a} n $$

In effect $\Delta x$ is the base of the rectangle, and $f(x)$ is the height. If we want to make it a triangle we insert $\frac 12$:

$$A = \lim \limits_{n \rightarrow \infty} \sum_{i=1}^{n} \frac 12 f(x_{i-1}) \Delta x$$

Since, we are summing triangles along the circumference, $b-a = 2 \pi r - 0$ and $\Delta x = \frac {2\pi r} n$. We can also conclude that $f(x)$ is a constant function equal to $\pi$, and we will need to insert to constant $\frac 1 2$.

$$A = \lim \limits_{n \rightarrow \infty} \sum_{i=1}^{n} \frac 12 r \frac {2\pi r} n$$

Pulling out the constants in regards to the summation and simplifying:

$$A = \lim \limits_{n \rightarrow \infty} \frac {\pi r^2} n \sum_{i=1}^{n} 1$$

Since $\sum_{i=1}^{n} 1 = n$:

$$A = \lim \limits_{n \rightarrow \infty} \frac {\pi r^2} n n = \pi r^2$$

(3) The above is probably the solution the problem wants. But we can take this one step further and convert the sum to an integral.

$$A = \lim \limits_{n \rightarrow \infty} \sum_{i=1}^{n} \frac 12 f(x_{i-1}) \frac {b-a} n = \frac 12 \int_{a}^{b}f(x)dx$$

Finally we have:

$$A = \frac 12 \int_{0}^{2\pi r}rdx = \frac 12 rx \Big|_{0}^{2\pi r} = \frac 12 r(2\pi r) - \frac 12 r(0) = \pi r^2$$


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