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Question Number 164626 by amin96 last updated on 19/Jan/22

Commented by mr W last updated on 19/Jan/22

$\Rightarrow C)$
	Answered by mr W last updated on 20/Jan/22

	Commented by mr W last updated on 20/Jan/22

	$s a y w e h a v e a r e g u l a r p o l y g o n w i t h$ $n s i d e s .$ $R = r a d i u s$ $a = s i d e l e n g t h$ $α = \frac{π}{n}$ $a = 2 R \sin α$ $l e t d_{i} = {P P}_{i}$ $a^{2} = d_{1}^{2} + d_{2}^{2} - 2 d_{1} d_{2} \cos α$ $a^{2} = d_{2}^{2} + d_{3}^{2} - 2 d_{2} d_{3} \cos α$ $. . .$ $a^{2} = d_{n - 1}^{2} + d_{n}^{2} - 2 d_{n - 1} d_{n} \cos α$ $a^{2} = d_{n}^{2} + d_{1}^{2} - 2 d_{n} d_{1} \cos (n - 1) α \leftarrow \cos (n - 1) α = \cos (π - α) = - \cos α$ $Σ :$ ${n a}^{2} = 2 (d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2}) - 2 (d_{1} d_{2} + d_{2} d_{3} + . . . + d_{n - 1} d_{n} - d_{n} d_{1}) \cos α$ ${n a}^{2} = 2 (d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2}) - \frac{1}{2} (d_{1} d_{2} + d_{2} d_{3} + . . . + d_{n - 1} d_{n} - d_{n} d_{1}) \sin α \times \frac{4}{\tan α}$ ${n a}^{2} = 2 (d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2}) - A_{p o l y g o n} \times \frac{4}{\tan α}$ $d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2} = \frac{n}{2} a^{2} + \frac{2 A_{p o l y g o n}}{\tan α}$ $A_{p o l y g o n} = n \times \frac{R^{2} \sin 2 α}{2} = \frac{{n R}^{2}}{2} \sin 2 α$ $d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2} = \frac{n}{2} \times 4 R^{2} \sin^{2} α + \frac{2 {n R}^{2} \sin 2 α}{2 \tan α}$ $d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2} = 2 {n R}^{2} \sin^{2} α + 2 {n R}^{2} \cos^{2} α$ $\Rightarrow d_{1}^{2} + d_{2}^{2} + . . . + d_{n}^{2} = 2 {n R}^{2}$ $g e n e r a l l y f o r r e g u l a r n - s i d e p o l y g o n :$ $\begin{matrix} {P P}_{1}^{2} + {P P}_{2}^{2} + . . . + {P P}_{n}^{2} = 2 {n R}^{2} \end{matrix}$ $i n c a s e o f o c t a g o n :$ $n = 8, R = 1$ ${P A}^{2} + {P B}^{2} + . . . + {P H}^{2} = 2 \times 8 \times 1^{2} = 16$
	Commented by Tawa11 last updated on 19/Jan/22

	$Weldone sir . Great$
	Commented by amin96 last updated on 20/Jan/22

	$bravo sir nice solution$
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