un paralelipiped dreptunghic ABCDA”B”C”D” are BD”=25cm si aria bazei egala cu 108 cm patrati
iar AB si BC sunt direct proportionale si 0,(3) si 0,25
aflati dimensiunile bazei
va roog ajutati-ma 
P.S vreau rezolvarea completa

Răspuns :

[tex] \dfrac{AB}{0,(3)}=\dfrac{BC}{0,25}\Rightarrow\dfrac{AB}{ \dfrac{1}{3} }=\dfrac{BC}{ \dfrac{1}{4} }\Rightarrow3AB=4BC\RightarrowAB=\frac{4}{3}BC[/tex]

Scriem acum aria
AB·BC=108⇒4/3·BC²=108⇒BC²=81⇒BC=9 cmAB=(4/3)·9=12 cm.
A=l*L=108 A=AB*BC AB BC dp cu 3/9=1/3 si 25/100 AB supra 1/3=Bc supra 1/4=k AB=k/3 si BC=k/4 si inlocuim in formula ariei 108=k/3*k/4 k la a2a supra 12=108 k la a2a=12*108 k la a2a=1296 k=v1296 v-radical k=36 AB-k/3=36/3=12 BC-k/4=36/4=9