Let $A_{1} A_{2} A_{3} \ldots A_{9}$ be a nine-sided regular polygon with side length 2 units. The difference between the lengths of the diagonals $\mathrm{A}_{1} \mathrm{~A}_{5}$ and $\mathrm{A}_{2} \mathrm{~A}_{4}$ equals

• A

  $2+\sqrt{12}$
• B

  $\sqrt{12}-2$
• C

  6
• Correct answer
  D

  2

Solution

$\angle \mathrm{A}_{1} \mathrm{OA}_{2}=\frac{2 \pi}{9}=40^{\circ}$ $\cos \frac{2 \pi}{9}=\frac{x^{2}+x^{2}-4}{2 x^{2}}$ $\left.x^{2} \cos 40^{\circ}\right)=x^{2}-2$ $x^{2}\left(1-\cos 40^{\circ}\right)=2$ $x=\frac{1}{\sin 20^{\circ}}$ .........(i) Now $\cos \frac{8 \pi}{9}=\frac{x^{2}+x^{2}-\left(A_{1} A_{5}\right)^{2}}{2 x^{2}}$ (in $\left.\Delta A_{1} O A_{5}\right)$ $\begin{array}{l} \left(A_{1} A_{5}\right)^{2}=2 x^{2}\left(1-\cos \frac{8 \pi}{9}\right) \\ =2 x^{2}\left(1-\cos 160^{\circ}\right) \\ =4 x^{2} \sin ^{2} 80^{\circ} \end{array}$ $\mathrm{A}_{1} \mathrm{~A}_{5}=2 \mathrm{x} \sin 80^{\circ}$ .........(ii) Similarly in $\triangle \mathrm{A}_{2} \mathrm{OA}_{4}$ $\mathrm{~A}_{2} \mathrm{~A}_{4}=2 \mathrm{x} \sin 40^{\circ}$ .........(iii) (ii) $-$ (iii) $\begin{aligned} \mathrm{A}_{1} \mathrm{~A}_{5}-\mathrm{A}_{2} \mathrm{~A}_{4} &=2 \mathrm{x}\left(\sin 80^{\circ}-\sin 40^{\circ}\right) \\ &=2(\text { using }(\mathrm{i})) \end{aligned}$

Hence, the correct answer is option: 'D'.