Trace as alturas BQ=h e CP=h
[tex3]\mathsf{2S=(a+b)h (a ~e~ b ~são~ bases)\\
\triangle ABQ \sim MNF \\
\triangle MEN \sim \triangle CPD\\
T.~ base~ média : MN=\frac{a+b}{2}=\frac{S}{h}\\
\frac{2MB}{\frac{S}{h}}=\frac{h}{FN}(I)\\
Analogamente:\\
\frac{2CN}{\frac{S}{h}} =\frac{h}{ME}(II)\\
(I)\cdot(II):\\
4MB*CN=\frac{(S^2*h^2)}{(h^2*FN*ME)}\implies
S^2=4MB*CN*ME*FN = 4*12*18\\
\therefore \boxed{\color{red}S=12\sqrt6}}
[/tex3]
(Solução:jvmago -
viewtopic.php?f=2&t=74529&p=202749&hili ... CD#p202749)