Defining regular polygons - class-IX

The side length is 2. The center of the hexagon is found by rotating the vector (A2-A1) by 60 degrees. With A1=(1,0) and A2=(3,0), the center is (2, sqrt(3)). The radius is 2. The circle equation is (x-2)^2 + (y-sqrt(3))^2 = 4, which simplifies to x^2 + y^2 - 4x - 2sqrt(3)y + 17/3 = 0.

Given ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ is  a regular hexagon inscribed in a circle of unit radius
$\Rightarrow {A} _{0}{A} _{1}=1$
$\Rightarrow {A} _{0}{A} _{2}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\Rightarrow {A} _{0}{A} _{4}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore {A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}=1\times \sqrt{3}\times\sqrt{3}=3$

Area of the regular polygon will be 
$=\dfrac{nR^{2}}{2}sin(\dfrac{2\pi}{n})$.
Now 
$R=\dfrac{s}{2sin(\dfrac{\pi}{n})}$
Hence
$A=\dfrac{ns^{2}}{8sin^{2}\dfrac{\pi}{n}}.2sin(\dfrac{\pi}{n}).cos(\dfrac{\pi}{n})$

$=\dfrac{ns^{2}}{4}.cot(\dfrac{\pi}{n})$. where s is the side of the polygon.

$=nr^{2}.tan(\dfrac{\pi}{n})$ where r is the incentre.

in $\Delta{ABL}$, $AL$ is the radius of the inscribed circle. 

The area of a regular hexagon with side length 'a' is (3*sqrt(3)/2) * a^2. For a=1, Area = 1.5 * 1.732 = 2.598 cm^2.

Let $a$ be the radius of the circle 

In a regular hexagon ABCDEF, the vector CE can be found using vector addition. Since AB = a and BC = b, the vectors for the sides are related by the geometry of the hexagon, leading to CE = b - a.

The area of a regular polygon with n sides and circumradius r is given by n * (1/2 * r^2 * sin(360/n)). This formula is derived by summing the areas of n isosceles triangles with sides r and included angle 360/n.

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

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 $86.5 m^2$.

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

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 $r=1,n=7$

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 $n=5$

$r + R = \dfrac {a}{2}\cot \dfrac {\pi}{n} + \dfrac {a}{2}cosec \dfrac {\pi}{n}$
$= \dfrac {a}{2} \left (\dfrac {1 + \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right ) = \dfrac {a}{2} \dfrac {2\cos^{2} \dfrac {\pi}{2n}}{2\sin \dfrac {\pi}{2n}\cdot \cos \dfrac {\pi}{2n}}$
$= \dfrac {a}{2} \cot \dfrac {\pi}{2n}$.

From the figure:
Side of polygon $(AB)=1$
$AO=\dfrac { 1 }{ 2 } $
$\angle O=\dfrac { \pi  }{ 2n } $

In right angled $\triangle COA$ :
$\sin { O } =\dfrac { AC }{ AO } $
$\Rightarrow \sin { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2R } $       ..(1)

$\tan { O } =\dfrac { AC }{ CO } $
$\Rightarrow \tan { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2r } $       ...(2)

From (1) and (2)
$R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ \sin { \dfrac { \pi  }{ n }  }  } +\dfrac { 1 }{ \tan { \dfrac { \pi  }{ n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1+\cos { \dfrac { \pi  }{ n }  }  }{ \sin { \dfrac { \pi  }{ n }  }  }  \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 2\cos ^{ 2 }{ \dfrac { \pi  }{ 2n }  }  }{ 2\cos { \dfrac { \pi  }{ 2n }  } \sin { \dfrac { \pi  }{ 2n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \cot { \dfrac { \pi  }{ 2n }  } $

Ans: A