Răspuns:
Explicație pas cu pas:
[tex]V=Aria_{bazei}*h;~Aria_{bazei}=\frac{AB^{2}\sqrt{3} }{4} ,~deci~\frac{AB^{2}\sqrt{3} }{4} *AA^{'}=\frac{81}{2} \\Aria_{bazei}=Aria(ABB^{'}A^{'},~deci~\frac{AB^{2}\sqrt{3} }{4}=AB*AA^{'} |:AB,~AA^{'}=\frac{AB\sqrt{3} }{4},~inlocuim,~\frac{AB^{2}\sqrt{3} }{4} *\frac{AB\sqrt{3} }{4}=\frac{81}{2},~\frac{AB^{3}*3}{16}=\frac{81}{2},~deci~AB^{3}=\frac{81*16}{2*3}=27*8=(3*2)^{3},~deci~AB =6cm\\ AA^{'}=\frac{AB\sqrt{3} }{4}=\frac{6\sqrt{3} }{4}=\frac{3\sqrt{3} }{2}.\\[/tex]
a) Aria(ΔC'AB)=???
C'A=C'B ca diagonale a dreptunghiurilor congruente, deci ΔC'AB isoscel cu baza AB. Fie C'D mediana in ΔC'AB, ⇒C'D⊥AB. Aflam CD din ΔCBD,
CD²=CB²-BD²=6²-3²=3²·2²-3²=3²·(2²-1)=3²·3. Deci CD=3√3cm.
CD⊥AB, ⇒C'D⊥AB, dupa T3⊥. Atunci C'D²=CD²+C'C²=(3√3)²+(3√3/2)²=3²·3+3²·3/4=3²·(3+3/4)=3²·15/4, deci C'D=(3/2)·√15.
Atunci Aria(ΔC'AB)=(1/2)·AB·C'D=(1/2)·6·(3/2)·√15=(9/2)·√15=4,5√15cm².
b) Aria(laterala)=Perimetrul(bazei)·h=3·AB·A'A=3·6·(3√3)/2=27√3cm².
