All related to triangles Part 3

Let $(a=\sqrt{13})$ , $(b=\sqrt{61})$ , $(c=\sqrt80)$

According to the question, we need to find: $\sqrt{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)}\\ =\sqrt{(a^{2}+b^{2}-c^{2}+2ab)×(-a^{2}-b^{2}+c^{2}+2ab)}$

Substituting the values of $a$ , $b$ and $c$ in the above expression, we get,

$\sqrt{(13+61-80+2×\sqrt{13}×\sqrt{61})×(-13-61+80+ 2×\sqrt{13}×\sqrt{61})} \\ =\sqrt{ (6+2×\sqrt{13}×\sqrt{61}×)(-6+2×\sqrt{13}×\sqrt{61})}$

Now, using $(a+b)×(a-b)=(a^{2}-b^{2})$ formula in the above expression, we get,

$\sqrt{(2×\sqrt{13}×\sqrt{61})^{2}-6^{2}}\\=\sqrt{(4×13×61)-36}\\ = \sqrt{3172-36 }\\ =\sqrt{3136} \\=\boxed{56}$ $\huge\ddot\smile$

Here's another approach, though I posted a comment, here's a modified one(as suggested by @Pi Han Goh )

Let, $a = \sqrt{13}, b=\sqrt{61}, c=\sqrt{80}$

Therefore we need to find, $\sqrt{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)}$

From the previous questions we find that, area of a triangles with sides, $a = \sqrt{13}, b=\sqrt{61}, c=\sqrt{80}$ was $14$ ,

Hence, $\sqrt{\dfrac{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)}{16}}=14$

Now, multiplying the equation by $4$ we get,

$\sqrt{\dfrac{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)}{16}} \times 4=56$

$\sqrt{\dfrac{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)\times 16}{16}}=56$

Finally, we can say that, $\sqrt{(a+b+c)×(a+b-c)×(a-b+c)×(-a+b+c)}=56$

From the Challenge Master's hints. It is Heron's formula, where the area of a triangle is given by:

$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ , where $s = \sqrt{13}+\sqrt{61}+\sqrt{80}$ , $a = 2\sqrt{13}$ , $b = 2\sqrt{61}$ and $c = 2\sqrt{80}$ .

And that:

$-16\Delta^2 = \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & c^2 & b^2 \\ 1 & c^2 & 0 & a^2 \\ 1 & b^2 & a^2 & 0 \end{vmatrix} \quad \Rightarrow \Delta^2 = - \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 80 & 61 \\ 1 & 80 & 0 & 13 \\ 1 & 61 & 13 & 0 \end{vmatrix} = 3136$

$\Rightarrow \Delta = \boxed{56}$