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The sides lengths of a triangle $a,$ $b$ and $c$ verify: $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=12b.$$ The task is to find the area of the triangle.

I'm trying to apply the heron's formula: $$\dfrac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}} {4}.$$ How do i get to heron's formula from what i know? Or it can be solved in another way?

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I think in you equation ,must be $24b$ $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b\\ \sqrt{a-24} + b^2 -24b+144+|c-12\sqrt{3} |=0\\ \sqrt{a-24} + (b-12)^2+|c-12\sqrt{3} |=0 \\\underbrace{\sqrt{a-24}}_{\geq 0} +\underbrace{(b-12)^2}_{\geq 0} + \underbrace{|c-12\sqrt{3} |}_{\geq 0} =0$$ some of positive expressions are zero , so all af them are zero at the same time $$\begin{cases}\sqrt{a-24}=0 & a =24\\(b-12)^2=0 & b=12\\|c-12\sqrt{3} |=0 & c=12\sqrt3\end{cases}$$ now plug into Heron's formula

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  • $\begingroup$ But $(b-12)^2=b^2-24b+144$. Or am I wrong? $\endgroup$ – Michael M. Sep 21 '17 at 7:10
  • $\begingroup$ @MichaelM. :I am almost sure that ,there is $24b $ Just like you .If not ,we can't say anything more... .WE think the same thing happening. $\endgroup$ – Khosrotash Sep 21 '17 at 7:12
  • $\begingroup$ Thanks for the help. $\endgroup$ – Michael M. Sep 21 '17 at 7:13
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$$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144-12b=\sqrt{a-24} + (b-6)^2 +|c-12\sqrt{3} |+108>0.$$

Thus, we have no any triangle with your given.

If you mean $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b$$ then we get $$(a,b,c)=(24,12,12\sqrt3)$$ and now you can use the Heron's formula.

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