= Solution
For eutectoid <pearlite>, apply the <lever rule> between <ferrite> and <cementite> at the transformation <temperature>. Using standard approximate Fe–C values $C_e=0.76$, $C_\alpha=0.022$ and $C_{\rm cem}=6.67$ in weight percent carbon,
$$
\boxed{w_{\rm cem}=\frac{C_e-C_\alpha}{C_{\rm cem}-C_\alpha}
=\frac{0.76-0.022}{6.67-0.022}\simeq0.111,
\qquad w_\alpha\simeq0.889.}
$$
These are \b[mass fractions]. To compare with an unbiased geometric line fraction, convert using the phase densities:
$$
f_{\rm cem}=\frac{w_{\rm cem}/\rho_{\rm cem}}
{w_{\rm cem}/\rho_{\rm cem}+(1-w_{\rm cem})/\rho_\alpha}.
$$
With representative densities $\rho_{\rm cem}\simeq7.69$ and $\rho_\alpha\simeq7.87\,\mathrm{Mg\,m^{-3}}$, this gives about \b[$11.3\,\mathrm{vol\%}$ <cementite> and $88.7\,\mathrm{vol\%}$ <ferrite>]. Because the densities are similar, the mass and volume fractions differ only modestly. The phase-diagram values here are conventional approximations, not a claim that the exact 2018 Data Book has been recovered.
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