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Answer:
[tex]\boxed{Circuit_{\Delta ABC} = 15~cm }[/tex]
Step-by-step explanation:
The ABC triangle is an equilateral triangle.
[tex]\mid AB\mid = \mid BC\mid = \mid AC\mid[/tex]
The drawing in the attachment.
I am entering meaningfully:
a - side length of an equilateral triangle
[tex]a=\mid AB\mid = \mid BC\mid = \mid AC\mid[/tex]
[tex]P_{\Delta} =\dfrac{a^{2} \sqrt{3} }{4}[/tex] - equilateral triangle area formula
[tex]P_{\Delta} =\dfrac{25 \sqrt{3} }{4}~~\land~~P_{\Delta} =\dfrac{a^{2} \sqrt{3} }{4}~~\Rightarrow~~\dfrac{a^{2} \sqrt{3} }{4}=\dfrac{25 \sqrt{3} }{4}\\\\\\\dfrac{a^{2} \sqrt{3} }{4}=\dfrac{25 \sqrt{3} }{4}~~\mid~~\div ~~\dfrac{\sqrt{3} }{4} \\\\\\\dfrac{a^{2} \sqrt{3} }{4}\cdot \dfrac{4}{\sqrt{3} } =\dfrac{25 \sqrt{3} }{4}\cdot \dfrac{4}{\sqrt{3} } \\\\a^{2} =25~~\\\\a^{2} =5^{2} ~~\land~~a > 0~~\Rightarrow~~\boxed{a=5~cm}\\\\\\[/tex]
[tex]Circuit_{\Delta ABC} =a+a+a\\\\Circuit_{\Delta ABC} =3a~~\land~~a=5~cm\\\\\boxed{Circuit_{\Delta ABC} =3\cdot 5 = 15~cm }[/tex]
