Perimeter from medians

From Apollonius' theorem:

$m_a = \sqrt{\frac{2b^2 + 2c^2 - a^2}{4}}$

$m_b = \sqrt{\frac{2a^2 + 2c^2 - b^2}{4}}$

$m_c = \sqrt{\frac{2a^2 + 2b^2 - c^2}{4}}$

which means:

$a = \frac{2}{3}\sqrt{2m_b^2 + 2m_c^2 - m_a^2} = \frac{2}{3}\sqrt{2 \cdot 6^2 + 2 \cdot 7^2 - 5^2} = \frac{2}{3}\sqrt{145}$

$b = \frac{2}{3}\sqrt{2m_a^2 + 2m_c^2 - m_b^2} = \frac{2}{3}\sqrt{2 \cdot 5^2 + 2 \cdot 7^2 - 6^2} = \frac{2}{3}\sqrt{112}$

$c = \frac{2}{3}\sqrt{2m_a^2 + 2m_b^2 - m_c^2} = \frac{2}{3}\sqrt{2 \cdot 5^2 + 2 \cdot 6^2 - 7^2} = \frac{2}{3}\sqrt{73}$

so the perimeter is $a + b + c = \frac{2}{3}\sqrt{145} + \frac{2}{3}\sqrt{112} + \frac{2}{3}\sqrt{73} \approx \boxed{20.77}$ .

Let the lengths of the sides of the triangle be $a, b, c$ . Then, using Apollonius' theorem, we get

$b^2+c^2=2\times 5^2+2\times \frac{a^2}{4}$

$c^2+a^2=2\times 7^2+2\times \frac{b^2}{4}$

$a^2+b^2=2\times 6^2+2\times \frac{c^2}{4}$

So, $a^2+b^2+c^2=\frac{4}{3}(5^2+7^2+6^2)=\frac{440}{3}$

So, $a=\frac{2}{3}\sqrt {145},b=\frac{2}{3}\sqrt {73}, c=\frac{2}{3}\sqrt {112}$ .

So, the perimeter of the triangle is

$\frac{2}{3}(\sqrt {145}+\sqrt {73}+\sqrt {112})\approx \boxed {20.779}$ .