Tag: regular polygons

For a regular hexagon with apothem $5 m$ the side length is about $5.77m$.
Use the formula
$A = \frac{1}{2}pa$
to find the area of the hexagon.
The perimeter of the hexagon is about $6(5.77)$ or $34.62m$.
Now substitute the values.
$A = \frac{1}{2} (34.62)(5)$
Simplify.
$A = \frac{1}{2}(173.1)$
= $86.5$
$A = \frac{1}{2}(173.1)$
= $86.5$
Therefore, the area of the regular hexagon is about $86.5 m^2$.

In $\triangle ACD, \cos{C}=\dfrac{b}{\left(\dfrac{a}{2}\right)}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{2b}{a}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2b}=2b$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=4{b}^{2}$
$\Rightarrow {a}^{2}-{c}^{2}=3{b}^{2}$

Let $A=2\alpha, B=3\alpha, C=7\alpha$
$\therefore A+B+C={180}^{0}$
$\Rightarrow 2\alpha+3\alpha+7\alpha={180}^{0}$
$\Rightarrow 12\alpha={180}^{0}$
$\Rightarrow \alpha={15}^{0}$
$\therefore A=2\alpha=2\times{15}^{0}={30}^{0}$
$B=3\alpha=3\times{15}^{0}={45}^{0}$
$C=7\alpha=7\times{15}^{0}={105}^{0}$
$a:b:c=\sin{{30}^{0}}:\sin{{45}^{0}}:\sin{{105}^{0}}$
       $=\dfrac{1}{2}:\dfrac{1}{\sqrt{2}}:\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$a:b:c=\sqrt{2}:2:\sqrt{3}+1$
        $=\dfrac{1}{\sqrt{2}}:1:\dfrac{\sqrt{3}+1}{2}$
or $1:\sqrt{2}:\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}$
or $1:\sqrt{2}:\dfrac{2\sqrt{2}}{2\left(\sqrt{3}-1\right)}$
$\therefore a:b:c= 1:\sqrt{2}:\dfrac{\sqrt{2}}{\sqrt{3}-1}$

$\cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$
Using transformation angle formula, we have
$\left(\cos{A}+\cos{B}\right)+\cos{C}=\dfrac{3}{2}$
$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+\cos{C}=\dfrac{3}{2}$
Using sub-multiple angle formula to $\cos{C}=1-2{\sin}^{2}{\dfrac{C}{2}}$ we get
$\Rightarrow  2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+1-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}$ 
Since $A+B+C=\pi\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$
$\Rightarrow  2\cos{\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}-1$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{C}{2}\right)}\right]=\dfrac{1}{2}$
Again
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\cos\left(\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{C}{2}\right)}\left[2\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{A}{2}\right)}\right]=\dfrac{1}{4}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{8}$
$\therefore \sin{\left(\dfrac{A}{2}\right)}=\dfrac{1}{2}, \sin{\left(\dfrac{B}{2}\right)}=\dfrac{1}{2},\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{2}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}=\sin{\dfrac{\pi}{6}}, \sin{\left(\dfrac{B}{2}\right)}=\sin{\dfrac{\pi}{6}},\sin{\left(\dfrac{C}{2}\right)}=\sin{\dfrac{\pi}{6}}$
$\therefore {\left(\dfrac{A}{2}\right)}={\left(\dfrac{B}{2}\right)}={\left(\dfrac{C}{2}\right)}=\dfrac{\pi}{6}$
$\Rightarrow \angle{A}=\angle{B}=\angle{C}=2\times\dfrac{\pi}{6}=\dfrac{\pi}{3}$
Hence, the triangle is equilateral.

${A} _{0}{A} _{1}=2.1.\cos{\dfrac{\pi}{3}}=2.1.\dfrac{1}{2}=1={A} _{1}{A} _{2}$
$\cos{\dfrac{2\pi}{3}}=\dfrac{{\left({A} _{0}{A} _{1}\right)}^{2}+{\left({A} _{1}{A} _{2}\right)}^{2}-{\left({A} _{0}{A} _{2}\right)}^{2}}{2{A} _{0}{A} _{1}.{A} _{1}{A} _{2}}$
$\Rightarrow \cos{\left(\pi-\dfrac{\pi}{3}\right)}=\dfrac{1+1-{\left({A} _{0}{A} _{2}\right)}^{2}}{2.1.1}$
$\Rightarrow -\cos{\dfrac{\pi}{3}}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow \dfrac{-1}{2}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow -1=2-{\left({A} _{0}{A} _{2}\right)}^{2}$
$\Rightarrow {\left({A} _{0}{A} _{2}\right)}^{2}=3$
$\therefore {A} _{0}{A} _{2}=\sqrt{3}={A} _{0}{A} _{4}$
$\therefore {A} _{0}{A} _{1}\times {A} _{0}{A} _{2}\times{A} _{0}{A} _{4}=1\times\sqrt{3}\times\sqrt{3}=3$

Inscribed circle of a regular polygon of $n$ sides
${A} _{1}=n{r}^{2}\tan{\dfrac{\pi}{n}}$
Here $n=8$
$\therefore {A} _{1}=8{r}^{2}\tan{\dfrac{\pi}{8}}$
Circumscribed circle of a regular polygon of $n$ sides is
${A} _{2}=\dfrac{n{R}^{2}}{2}\sin{\dfrac{2\pi}{n}}$
For $n=8$ we have
${A} _{2}=\dfrac{8{R}^{2}}{2}\sin{\dfrac{2\pi}{8}}$
  $=\dfrac{8{R}^{2}}{2}\sin{\dfrac{\pi}{4}}$
  $=\dfrac{8{r}^{2}}{2}2\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ (for $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$

If radius of circle is $r$ then 
${A} _{1}{A} _{2}=2r\sin{\left(\dfrac{\pi}{n}\right)}$
${A} _{1}{A} _{3}=2r\sin{\left(\dfrac{2\pi}{n}\right)}$
${A} _{1}{A} _{4}=2r\sin{\left(\dfrac{3\pi}{n}\right)}$
$\because \dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}}$
$\Rightarrow \dfrac{1}{2r\sin{\left(\dfrac{\pi}{n}\right)}}=\dfrac{1}{2r\sin{\left(\dfrac{2\pi}{n}\right)}}+\dfrac{1}{2r\sin{\left(\dfrac{3\pi}{n}\right)}}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{3\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}+\sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\left[\sin{\left(\dfrac{3\pi}{n}\right)}-\sin{\left(\dfrac{\pi}{n}\right)}\right]=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}.2\cos{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow 2\sin{\left(\dfrac{2\pi}{n}\right)}\cos{\left(\dfrac{2\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
Using multiple angle formula, $2\sin{A}\cos{A}=\sin{2A}$ we get
$\sin{\left(\dfrac{4\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
$\therefore \dfrac{4\pi}{n}=r+{\left(-1\right)}^{r}\dfrac{3}{n}$ for $r=1,n=7$

$\because b+c=x,bc=y$
$\therefore b,c$ are the roots of ${t}^{2}-\left(b+c\right)t+bc=0$
or ${t}^{2}-xt+y=0$
$\therefore t=\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2}$
Hence, sides are  $\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2},z$ 

Let $a$ be the length of side of regular polygon then
$R=\dfrac{a}{2}\csc{\left(\dfrac{\pi}{n}\right)}$ and $r=\dfrac{a}{2}\cot{\left(\dfrac{\pi}{n}\right)}$
$\therefore \dfrac{R}{r}=\dfrac{\csc{\left(\dfrac{\pi}{n}\right)}}{\cot{\left(\dfrac{\pi}{n}\right)}}=\dfrac{\dfrac{1}{\sin\left(\frac{\pi}{n}\right)}}{\dfrac{\cos{\frac{\pi}{n}}}{\sin{\frac{\pi}{n}}}}=\dfrac{1}{\cos{\left(\frac{\pi}{n}\right)}}$
$\therefore \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}$
On rationalising the denominator, we get 
$ \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1}{4}=\cos{\left(\dfrac{\pi}{5}\right)}$
On comparing $\cos{\left(\dfrac{\pi}{n}\right)}=\cos{\left(\dfrac{\pi}{5}\right)}$ we get $n=5$