Skip to content

DA2. Polygon and Circle

Statement

  • If an n-sided regular polygon is inscribed in a circle of radius r, as shown in the figure below, then n-isosceles triangles fill the circle.
  • Based on the statement and figure above answer the following:
    1. Express h and the base b of the isosceles triangle shown in terms of θ and r.
    2. Express the area of the isosceles triangle in terms of θ and r. Use trig identities as needed.
    3. Describe what happens asn goes to infinity, (notice the polygon fills the circle, the angle θ goes to zero)
    4. Use special limit rules to discuss your response, you may use any graphing tool to support your response.

Solution

1. Express h and the base b of the isosceles triangle shown in terms of θ and r

According to (TestBook, 2023), The altitude of an isosceles triangle bisects the apex angle and also bisects the base. Thus inside one of the isosceles triangles, we have two congruent right triangles. With each of them having a hypotenuse of length \(r\) and an angle of \(\frac{θ}{2}\), two right sides are equal to \(\frac{b}{2}\) and \(h\). Therefore, we have:

\[ \begin{aligned} \sin\left(\frac{θ}{2}\right) &= \frac{b}{2r} \\ b &= 2r\sin\left(\frac{θ}{2}\right) \\ \cos\left(\frac{θ}{2}\right) &= \frac{h}{r} \\ h &= r\cos\left(\frac{θ}{2}\right) \end{aligned} \]

2. Express the area of the isosceles triangle in terms of θ and r. Use trig identities as needed

The area of the isosceles triangle is: \(A = \frac{1}{2}bh\). Therefore, where \(b\) is the base and \(h\) is the height of the triangle, we have:

\[ \begin{aligned} A &= \frac{1}{2}bh \\ &= \frac{1}{2}\left(2r\sin\left(\frac{θ}{2}\right)\right)\left(r\cos\left(\frac{θ}{2}\right)\right) \\ &= r^2\sin\left(\frac{θ}{2}\right)\cos\left(\frac{θ}{2}\right) \\ &= r^2( \frac{1}{2} (\sin(\frac{θ}{2} + \frac{θ}{2}) + \sin(\frac{θ}{2} - \frac{θ}{2})) ) \text{ ...... product-to-sum} \\ &= \frac{r^2}{2}(\sin(θ) + \sin(0)) \\ &= \frac{r^2}{2}\sin(θ) \end{aligned} \]

The above equation gives us the area of one isosceles triangle. Since there are \(n\) isosceles triangles in the polygon, the area of the polygon is:

\[ \begin{aligned} A &= n\left(\frac{r^2}{2}\sin(θ)\right) \\ &= \frac{nr^2}{2}\sin(θ) \end{aligned} \]

3. Describe what happens asn goes to infinity, (notice the polygon fills the circle, the angle θ goes to zero)

  • When \(n\) goes to infinity, the angle \(\theta\) goes to zero, thus the area of each triangle goes to zero.
  • As a result, the base of each triangle \(b=2r\sin\left(\frac{θ}{2}\right)\) also go to zero.
  • Also, the height of each triangle \(h=r\cos\left(\frac{θ}{2}\right)\) also go to \(r\).
  • So the polygon now holds an infinite number of triangles with bases go to zero and heights go to \(r\).
  • Thus, the polygon fills the circle.

4. Use special limit rules to discuss your response, you may use any graphing tool to support your response

  • The limit of \(\sin(x)\) is zero as \(x\) goes to zero:
    • Thus the limit of \(\sin\left(\frac{θ}{2}\right)\) is zero as \(\theta\) goes to zero.
    • As a result, the limit of \(b=2r\sin\left(\frac{θ}{2}\right)\) goes to zero as \(\theta\) goes to zero.
  • The limit of \(\cos(x)\) is one as \(x\) goes to zero:
    • Thus the limit of \(\cos\left(\frac{θ}{2}\right)\) is one as \(\theta\) goes to zero.
    • As a result, the limit of \(h=r\cos\left(\frac{θ}{2}\right)\) goes to \(r.1=r\) as \(\theta\) goes to zero.

References