Get to the root of the problem...

$8-4\sqrt{3}=6-2\times 2\sqrt{3}+2=(\sqrt{6}-\sqrt{2})^{2}$ . So $a=6$ and $b=2$

$22a+10b=22\times 6+10\times 2=152$

I couldn't understand how it linked to geometry.....

Hint: Proceed with $8-4\sqrt{3}\equiv(\sqrt{a} - \sqrt{b})^2$ to get a+b = 8 and ab =12 (for a,b > 0) where the only solution is a =6 and b=2 which can be written as $\sqrt{6} - \sqrt{2}$ . This links to geometry because $\sqrt{6} - \sqrt{2} = 4sin15$ .