[tex3]\mathsf{\triangle ABH: (AB)^2 = (2n)^2+(2h)^2 \implies AB = 2\sqrt{(n^2+h^2)}\\
I+S_{ABC} = 40(i)\\
II+10 = \frac{\pi r^2}{2}(ii)\\
I+II+S_{ABC}=\frac{\pi R^2}{2}(iii)\\
(i)+(ii) = I+II+S_{ABC} = 30+\frac{\pi r^2}{2}\implies\\
De(iii): \frac{\pi (\underbrace{R^2- r^2})}{2} = 30(iv)\\
x+ y = \frac{\pi n^2}{2}+\frac{\pi h^2}{2}=\frac{\pi}{2}(n^2+h^2)(V)\\
OD = \frac{AB}{2} =\sqrt{n^2+h^2} \\\triangle ODC: n^2+h^2+r^2 = R^2 \implies \underbrace{R^2-r^2}=n^2+h^2\\
\therefore De(iv)e (v): \boxed{\color{red}\frac{\pi (n^2+h^2)}{2} = 30=x+y}
}[/tex3]
(Solução: Null -
viewtopic.php?t=89401)