Question

Intr-o piramida triunghiulara regulata, raza cercului circumscris bazei este de 6 radical din 3 cm iar apotema piramidei este de 3 radical din 7 cm. Calculati:
a) muchia laterala piramidei
b) volumul piramidei

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it OM=\dfrac{OC}{2}=\dfrac{6\sqrt3}{2}=3\sqrt3\ cm\\ \\ \\ CM= OC+OM=6\sqrt3+3\sqrt3=9\sqrt3\ cm\\ \\OC=\dfrac{\ell\sqrt3}{2}=9\sqrt3 \Rightarrow\ \ell=18\ cm$

$\it MB=\dfrac{\ell}{2}=\dfrac{18}{2}=9\ cm\\ \\ \Delta VBB-dreptunghic,\ m(\hat M)=90^o\ \stackrel{Th.\ Pitagora}{\Longrightarrow}\ VB^2=VM^2+MB^2\ \Rightarrow \\ \\ \Rightarrow\ VB^2=(3\sqrt7)^2 +9^2=63+81=144=12^2\ \Rightarrow\ VB=12\ cm$

$\it \mathcal{V}=\dfrac{\mathcal{A}_b\ \cdot h}{3};\ \ \ \mathcal{A}_b=\dfrac{\ell^2\sqrt3}{4}=\dfrac{18\cdot18\sqrt3}{4}=81\sqrt3\ cm^2$

$\it\ \Delta VOM-dreptunghic,\ m(\hat O)=90^o,\ \stackrel{Th.Pitagora}{\Longrightarrow}\ h^2=(3\sqrt7)^2-(3\sqrt3)^2=\\ \\ =63-27=36=6^2\ \Rightarrow\ h=6\ cm$