Question

Uma pirâmide regular hexagonal de 12 cm de altura tem aresta da base medindo 10√3

3

cm. Calcule:

a) o apótema da base b) o apótema da pirâmide (g) c) a aresta lateral d) a área da base (Ab)

e) a área lateral (Al) f) a área total (At) g) o volume (V)​

Answer 1

$\boxed{\begin{array}{l}\tt a)~\sf m=\dfrac{\diagdown\!\!\!\!\!\!1\!0\sqrt{3}\cdot\sqrt{3}}{\diagdown\!\!\!2}=5\cdot3=15~cm\\\tt b)~\sf a_p^2=h^2+m^2\\\sf a_p^2=12^2+15^2\\\sf a_p^2=144+225\\\sf a_p^2=369\\\sf a_p=\sqrt{369}=3\sqrt{41}~cm\end{array}}$

$\boxed{\begin{array}{l}\tt c)~\sf \ell^2=a_p^2+\bigg(\dfrac{10\sqrt{3}}{2}\bigg)^2\\\sf \ell^2=369+75\\\sf \ell^2=444\\\sf \ell=\sqrt{444}=2\sqrt{111}~cm\\\tt d)~\sf A_b=\dfrac{3\cdot(10\sqrt{3})^2\cdot\sqrt{3}}{2}\\\\\sf A_b=\dfrac{3\cdot300\cdot\sqrt{3}}{2}=450\sqrt{3}~cm^2\end{array}}$

$\boxed{\begin{array}{l}\tt e)~\sf A_l=\backslash\!\!\!6\cdot\dfrac{1}{\backslash\!\!\!2}\cdot10\sqrt{3}\cdot3\sqrt{41}\\\sf A_l=90\sqrt{123}~cm^2\\\tt f)~\sf A_t=A_l+A_b\\\sf A_t=90\sqrt{123}+450\sqrt{3}\\\sf A_t=90\sqrt{3}(\sqrt{41}+5)~cm^2\end{array}}$

$\large\boxed{\begin{array}{l}\tt g)~\sf V=\dfrac{1}{3}\cdot A_b\cdot h\\\\\sf V=\dfrac{1}{\diagdown\!\!\!\!3}\cdot450\sqrt{3}\cdot\diagdown\!\!\!\!\!\!1\!2\\\sf V=4\cdot450\sqrt{3}=1800\sqrt{3}~cm^3\end{array}}$