Let a, b, c be the lengths of sides of triangle ABC such that \(\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}\) = k. Then \( \frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}\) = 

1
36
2
32
3
38
4
40

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