General Polygons - Problem Solving

We shall first study regular polygons. A regular polygon is a polygon in which all sides have the same length and all angles are equal in measure. A rhombus is not a regular polygon, though all sides are equal. This is because all angles are not equal. Let \(S\) be the side length of an \(n\)-sided polygon (\(n\)-gon), \(R\) the circumradius, and \(r\) the inradius.

An \(n\)-gon is made up of \(n\) isosceles triangles with base \(S,\) sides \(R,\) and angle between the two sides measuring \(\frac{2\pi} n.\)

For relations between various lengths and angles, the right-angled triangle, which is half of the isosceles triangle, is helpful. So we get the relation \[S=2\times R\sin\frac \pi n ~\text{ or }~ R=\dfrac S{2\times \sin\frac \pi n }\] and \[S=2\times r\tan\frac \pi n ~\text{ or }~ r=\dfrac {S\times \cot\frac \pi n } 2. \] The base angles are \(\frac{(n-2)\times \pi}{2n},\) while the angle between adjacent sides is \(\frac{(n-2)\times\pi}{n}.\)

The sketch below shows how the isosceles triangles are placed with a common vertex, coinciding sides, and the angles. Apart from the solution of the Isosceles triangle, the bases of \(n\)-gon solution, the lengths of diagonals \(D_2, D_3, D_4, \ldots\) are important.

For an \(n\)-gon, there will be \(n\) vertices \(A_0, A_1, \ldots, A_{n-2}, A_{n-1}. \) Then \((n-1)\) lines can be drawn from \(A_0\) to othere vertices. Out of these, \(A_0A_1\) and \(A_0A_{n-1} \) are adjacent sides. So there are only \((n-3)\) diagonals. If \(n\) is even, there will be one big diagonal, and all the \((n-2)\) remaining ones will be in \(\frac{n-4} 2\) pairs, as shown in the sketch. If \(n\) is odd, all \((n-3)\) diagonals will be in \(\frac{n-3} 2\) pairs. What is true for vertex \(A_0\) is true for each of the \(n\) vertices.

Next, we have to find out the lengths of diagonals.

The diagonal \(D_n\) between \(A_0A_k\) also forms an isosceles triangle with sides \(R\) and the angle \(\alpha=\frac{2\pi} n\times k.\) Now we can solve these as we had solved for sides.

\[\dfrac{144}{5}+\dfrac{128}{25}\sqrt{3}\] \[\dfrac{24}{5}\sqrt{5}+\dfrac{64}{75}\sqrt{3}\] \[6\left(\dfrac{16}{15}\sqrt{3}+3\right)\] \[\sqrt{3}\left(\dfrac{144}{5}\sqrt{2}+\dfrac{128}{25}\right)\]

A regular hexagon swimming pool has a side length of \(3\) meters. It is surrounded by a sidewalk of uniform width \(1.2\) meters. The sidewalk is surrounded by a grass landscape of uniform width \(0.4\) meters.

Which of the following is the combined area of the sidewalk and landscape?