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Given, ratios of the sides as \[12:17:25\]

Perimeter of the triangle, \[P = 540\,{\text{cm}}\]

Let us first find the sides of the triangle.

Let ABC be the triangle, with the ratios of the sides as

\[{\text{AB}}:{\text{BC}}:{\text{CA}} = 12:17:25\]

Let \[x\] be a constant such that \[{\text{AB}}:{\text{BC}}:{\text{CA}} = 12x:17x:25x\]

Perimeter of a triangle is given by the sum of its sides, so perimeter of triangle ABC will be,

\[P = {\text{AB}} + {\text{BC}} + {\text{CA}}\]

Putting the values of \[P\] and the sides AB, BC and AC we get

\[

540\, = 12x + 17x + 25x \\

\Rightarrow 540 = 54x \\

\Rightarrow x = 10

\]

Therefore, the sides are

\[{\text{AB}} = 12x = 12 \times 10 = 120\,{\text{cm}}\]

\[{\text{BC}} = 17x = 17 \times 10 = 170\,{\text{cm}}\]

\[{\text{AC}} = 25x = 25 \times 10 = 250\,{\text{cm}}\]

Area of a triangle is given by,

\[{\text{Area}} = \sqrt {s(s - a)(s - b)(s - c)} \] (i)

where \[s\] is the semi perimeter and \[a\] , \[b\] and \[c\] are its sides.

Here,

\[

s = \dfrac{P}{2} \\

\Rightarrow s = \dfrac{{540}}{2} \\

\Rightarrow s = 270{\text{cm}}

\]

And \[a\] , \[b\] and \[c\] are

\[a = {\text{AB}} = 120{\text{cm}}\]

\[b = {\text{BC}} = 170{\text{cm}}\]

\[c = {\text{AC}} = 250{\text{cm}}\]

Putting these values of \[s\] , \[a\] , \[b\] and \[c\] in equation (i), we get

\[

{\text{Area}} = \sqrt {270(270 - 120)(270 - 170)(270 - 250)} \\

\Rightarrow {\text{Area}} = \sqrt {270 \times 150 \times 100 \times 20}

\]

\[

\Rightarrow {\text{Area}} = \sqrt {81000000} \\

\Rightarrow {\text{Area}} = 9000{\text{c}}{{\text{m}}^{\text{3}}}

\]

Thus, the area of the triangle is \[9000{\text{c}}{{\text{m}}^{\text{3}}}\] .

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