[tex] \frac{CD}{4}=x; \frac{AB}{6}=x
CD=4x;AB=6x \\
din.triunghiul ACD-dreptunghic,CH-inaltime \\ CH=AD=4 \sqrt{2}
\\ Teorema -inaltimiii \\ CH^2=AH*HB \\ 32=4x*2x \\ 32=8x \\ x=4 \\
deci,DC =4*4=16 cm \\ AB=6*4=24cm \\ A= \frac{b+B}{2}*h= \frac{16+24}{2}*4 \sqrt{2}=80 \sqrt{2}cm^2 [/tex]