Resistance of CD arm = 2r cos 72 o = 0.62r

Resistance of CBFC branch

$\frac{1}{R} = \frac{1}{2r} + \frac{1}{0.62r} = \frac{1}{r} \left( \frac{2.62}{2 \times 0.62} \right)$

$\frac{1}{R} = \frac{2.62}{1.24r}$ ∴ ∴ $R = \frac{1.24r}{2.62}$

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

∴ ∴ Equivalent resistance between A and H

$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'}$ ⇒ ⇒ $R_{eq} = \frac{R'}{2} = \frac{1.946}{2} r = 0.973 r$