Răspuns :
7)
[tex]\it L_c=2\pi R\ \Rightarrow\ R=\dfrac{L_c}{2\pi}=\dfrac{16\pi}{2\pi}=8\ cm\\ \\ \Delta AOB\ -\ isocel,\ \ OA=OB,\ \Rightarrow\ \hat A=\hat B=\dfrac{180^o-120^o}{2}=30^o\\ \\ \\ Th.\ sinusurilor\ \Rightarrow\ \dfrac{OB}{sinA}=\dfrac{AB}{sinO}\ \Rightarrow\ AB=\dfrac{OB\cdot sinO}{sinA}=\dfrac{8\cdot sin120^o}{sin30^o}=\\ \\ \\ =\dfrac{8\cdot\dfrac{\sqrt3}{2}}{\dfrac{1}{2}}=8\sqrt3\ cm[/tex]
9)
Deoarece avem piramidă regulată, muchiile laterale sunt congruente,
deci, fețele laterale sunt triunghiuri dreptunghice isoscele.
[tex]\it Fie\ VABC\ -\ piramida\ cu\ bnaza\ \Delta ABC\ -\ echilateral\\ \\ \mathcal{A}_{\ell}=3\cdot\mathcal{A}_{VAB}\ \Rightarrow\ \mathcal{A}_{VAB}=\dfrac{\mathcal{A}_{\ell}}{3}=\dfrac{12}{3}=4\ cm^2\ \ \ \ \ (1)\\ \\ \\ \mathcal{A}_{VAB}=\dfrac{VA\cdot VB}{2}=\dfrac{VA\cdot VA}{2}=\dfrac{VA^2}{2}\ \ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \dfrac{VA^2}{2}=4 \Rightarrow VA^2=2\cdot4=8 \Rightarrow VA=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\ cm[/tex]
[tex]\it \Delta VAB\ -\ dreptunghic\ isoscel\ \Rightarrow\ AB=VA\sqrt2=2\sqrt2\cdot\sqrt2=2\cdot2=4\ cm\\ \\ \mathcal{P}_{ABC}=3\cdot \ell=3\cdot4=12\ cm[/tex]