Question

Let $A_1{A}_2{A}_3....A_9$ be a nine-sided regular polygon with side length $2$ units. The difference between the lengths of the diagonals $A_1{A}_5$ and $A_2{A}_4$ equals

• A. $2+\sqrt{12}$
• B. $\sqrt{12}-2$
• C. $6$
• D. $2$

Answer 1

The correct option is D $2$

Given :
$A_1{A}_2{A}_3....A_9$ is a regular polygon

$A_1{A}_2=2$ units
Let $O{A}_1=x$
$\angle A_{1}O{A}_2=\frac{2\pi }{9}$
Applying cosine rule in $△A_{1}O{A}_2$,
$\cos \frac{2\pi }{9}=\frac{x^2+x^2-4}{2x^2}\Rightarrow \cos {40}^{\circ }=1-\frac{2}{x^2}\Rightarrow \frac{2}{x^2}=1-\cos {40}^{\circ }\Rightarrow \frac{2}{x^2}=2{\sin }^2{20}^{\circ }\Rightarrow x=\frac{1}{\sin {20}^{\circ }}...\left(i\right)$
Similarly using cosine rule in $△A_{2}O{A}_4$,
$\cos \frac{4\pi }{9}=\frac{x^2+x^2-(A_2{A}_4{)}^2}{2x^2}\Rightarrow 2(1-\cos {80}^{\circ })={\left(\frac{A_2{A}_4}{x}\right)}^2\Rightarrow \frac{A_2{A}_4}{x}=2\sin {40}^{\circ }$
$\Rightarrow A_2{A}_4=2x\sin {40}^{\circ }.....\left(ii\right)$

Similarly using cosine rule in $△A_{1}O{A}_5$,
$\cos \frac{8\pi }{9}=\frac{x^2+x^2-(A_1{A}_5{)}^2}{2x^2}\Rightarrow 2(1-\cos {160}^{\circ })={\left(\frac{A_1{A}_5}{x}\right)}^2\Rightarrow \frac{A_1{A}_5}{x}=2\sin {80}^{\circ }$
$\Rightarrow A_1{A}_5=2x\sin {80}^{\circ }.....\left(iii\right)$

$\therefore A_1{A}_5-A_2{A}_4=2x(\sin {80}^{\circ }-\sin {40}^{\circ })=2x\left(2\cos {60}^{\circ }\sin {20}^{\circ }\right)=2$