Explanation

(b) Resistance of $CD\ arm = 2r\cos{72{^\circ} = 0.62r}$
Resistance of CBFC branch
$\frac{1}{R} = \frac{1}{2r} + \frac{1}{0.62r} = \frac{1}{r}\left( \frac{2.62}{2 \times 0.62} \right)$
$\frac{1}{R} = \frac{2.62}{1.24r}\ \ \ \therefore R = \frac{1.24r}{2.62}$
Equivalent $R^{'} = 2R + r = 2 \times \frac{1.24r}{2.62} + r$
$= r\left( \frac{2.48}{2.62} + 1 \right) = 1.946r$
Because the star circuit is symmetrical about the line $AH$
$\therefore$ Equivalent resistance between $A$ and $H$
$\frac{1}{R_{eq}} = \frac{1}{R^{'}} + \frac{1}{R^{'}}$
$\Rightarrow R_{eq} = \frac{R^{'}}{2} = \frac{1.946}{2}r = 0.973r$