please.....bantuannnnnnnya

L. segitiga ABC adalah [tex] \left ( \frac{4\sqrt{2}}{2} \right )^2 \sqrt{3} = \left ( 2 \sqrt{2} \right )^2 \sqrt{3} = 8\sqrt{3}[/tex].
Maka luas daerah yang diarsir,
[tex]L_{arsir} = L_{\triangle{ABC}} - L_{\circ A} \\ \frac{8}{3} \left ( 3 \sqrt{3} - \pi \right ) = 8\sqrt{3} - \frac{1}{6} \pi r^2 \\ 8\sqrt{3} - \frac{8}{3} \pi = 8\sqrt{3} - \frac{1}{6} \pi r^2 \\ \frac{8}{3} \pi = \frac{1}{6} \pi r^2 \\ \frac{8}{3} = \frac{1}{6} r^2 \\ 16 = r^2 \\ \therefore r = 4[/tex]