Let a, b, c be the lengths of sides of triang | vedclass

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(A) In $	riangle ABC$,we have $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$.This gives:$a+b=7k$ $(i)$$b+c=8k$ $(ii)$$c+a=9k$ $(iii)$Adding these equa

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$36$

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(A) In $	riangle ABC$,we have $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$.This gives:$a+b=7k$ $(i)$$b+c=8k$ $(ii)$$c+a=9k$ $(iii)$Adding these equations,we get $2(a+b+c)=24k$,so $a+b+c=12k$ $(iv)$.Subtracting $(i), (ii), (iii)$ from $(iv)$ respectively:$c = (a+b+c) - (a+b) = 12k - 7k = 5k$$a = (a+b+c) - (b+c) = 12k - 8k = 4k$$b = (a+b+c) - (c+a) = 12k - 9k = 3k$Since $a^2+b^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 = (5k)^2 = c^2$,the triangle is a right-angled triangle with hypotenuse $c$.Thus,$\angle C = 90^{\circ}$.Area of $	riangle ABC = \frac{1}{2} 	imes a 	imes b = \frac{1}{2} 	imes 4k 	imes 3k = 6k^2$.Therefore,$\frac{(A(	riangle ABC))^2}{k^4} = \frac{(6k^2)^2}{k^4} = \frac{36k^4}{k^4} = 36$.

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 $\frac{(A(\triangle ABC))^2}{k^4}=$

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