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(A) Let the total resistance of the polygon be $R$. Since it is an $n$-sided polygon,the resistance of each side is $r = \frac{R}{n}$.When we consider

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$\frac{(n-1) R}{n^2}$

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(A) Let the total resistance of the polygon be $R$. Since it is an $n$-sided polygon,the resistance of each side is $r = \frac{R}{n}$.When we consider two adjacent corners,say $A$ and $B$,the circuit splits into two parallel paths:$1$. The direct side $AB$ with resistance $r$.$2$. The remaining $(n-1)$ sides connected in series with a total resistance of $(n-1)r$.The equivalent resistance $R_{eq}$ between $A$ and $B$ is given by the parallel combination of these two paths:$R_{eq} = \frac{r \cdot (n-1)r}{r + (n-1)r}$$R_{eq} = \frac{(n-1)r^2}{nr} = \frac{(n-1)r}{n}$Substituting $r = \frac{R}{n}$ into the equation:$R_{eq} = \frac{(n-1)}{n} \cdot \frac{R}{n} = \frac{(n-1)R}{n^2}$

The equivalent resistance between the adjacent corners of a regular $n$-sided polygon made of a uniform wire of total resistance $R$ is:

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