An unexpected approach to a quadratic system of equations

Positive reals $a,b,c$ satisfy

$$ a^2 + b^2  + ab = 121$$

$$b^2 + c^2 + bc = 169$$

$$ c^2 + a^2 + ca = 400$$

What is the value of $ab + bc + ca$?

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This can be solved using geometry!

Consider triangle $OAB$ with $OA = a$ and $OB = b$ and $\angle{AOB} = 120^{\circ}$.

Using cosine rule, $AB^2 = a^2 + b^2 - 2ab \cos(120^{\circ}) = a^2 + b^2 + ab$. Thus $AB = 11$.

Area of triangle $AOB$ = $\frac{1}{2} ab \sin(120^{\circ}) = \frac{\sqrt{3}}{4} ab$

Now add point $C$ such that $OC = c$ and $\angle{AOC} = \angle{COB} = 120^{\circ}$.

We thus have a triangle $ABC$ with sides $AB = 11, BC = 13, AC = 20$ and area  (by adding up areas of $AOB, AOC, BOC$

 $$\frac{\sqrt{3}}{4} (ab + bc + ca)$$

By Heron's formula, since the sides are known, we get area $=66$.

This gives $$ab + bc + ca = 88 \sqrt{3}$$