(i) The rate of growth of the infected area is proportional to the product of the infected area \(a\) and the uninfected area \(10 - a\). Therefore, \(\frac{da}{dt} = k a (10 - a)\). Given that initially \(a = 5\) and \(\frac{da}{dt} = 0.1\), we have:

\(0.1 = k \times 5 \times (10 - 5)\)

\(0.1 = 25k\)

\(k = 0.004\)

Thus, \(\frac{da}{dt} = 0.004a(10 - a)\).

(ii) Express \(\frac{1}{a(10-a)}\) in partial fractions:

\(\frac{1}{a(10-a)} = \frac{A}{a} + \frac{B}{10-a}\)

Solving, we find \(A = \frac{1}{10}\) and \(B = \frac{1}{10}\).

Separate variables and integrate:

\(\int \frac{da}{a(10-a)} = \int 0.004 \, dt\)

\(\frac{1}{10} \ln a - \frac{1}{10} \ln (10-a) = 0.004t + C\)

\(\ln \left( \frac{a}{10-a} \right) = 0.04t + C\)

Using initial conditions \(t = 0, a = 5\), solve for \(C\):

\(\ln 1 = C \Rightarrow C = 0\)

Thus, \(\ln \left( \frac{a}{10-a} \right) = 0.04t\)

\(t = 25 \ln \left( \frac{a}{10-a} \right)\)

(iii) For 90% infection, \(a = 9\):

\(t = 25 \ln \left( \frac{9}{1} \right)\)

\(t = 25 \ln 9 \approx 54.9\)