Answer
$A_{2}= 24.7$ cm$^{2} $
Work Step by Step
$$
\frac{A}{A^*}=\frac{1}{\mathrm{Ma}}\left\{\left(\frac{2}{k+1}\right)\left(1+\frac{k-1}{2} \mathrm{Ma}^2\right)\right\}^{(k+1) / 2(k-1)}
$$
For $k=1.33$ and $\mathrm{Ma}_1=1.8$ :
$$ \frac{A_1}{A^*}=\frac{1}{1.8}\left\{\left(\frac{2}{1.33+1}\right)\left(1+\frac{1.33-1}{2} 1.8^2\right)\right\}^{2.33 / 2 \times 0.33}=1.4696
$$ and,
$\quad A^*=\frac{A_1}{2.570}=\frac{36 \mathrm{~cm}^2}{1.4696}=24.50 \mathrm{~cm}^2$
For $k=1.33$ and $\mathrm{Ma}_2=0.9$: $$
\frac{A_2}{A^*}=\frac{1}{0.9}\left\{\left(\frac{2}{1.33+1}\right)\left(1+\frac{1.33-1}{2} 0.9^2\right)\right\}^{2.33 / 2 \times 0.33}=1.0091
$$ and
$\quad A_2=(1.0091) A^*=(1.0091)\left(24.50 \mathrm{~cm}^2\right)=\mathbf{2 4 . 7 \mathrm { cm } ^ { 2 }}$
