Question

Em uma pirâmide hexagonal regular, a área da base é 18 raiz de 3cm^2 e a área lateral, 30 raiz de 3cm^2. Obtenha seu volume.

Answer 1

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$\boxed{\begin{array}{l}\underline{\sf c\acute alculo~da~aresta~da~base\!:}\\\sf B=\dfrac{3a^2\sqrt{3}}{2}\\\sf 18\diagup\!\!\!\!\!\sqrt{3}=\dfrac{3a^2\diagup\!\!\!\!\!\sqrt{3}}{2}\\\sf a^2=\dfrac{2}{3}\cdot18\\\sf a^2=12\\\sf a=\sqrt{3\cdot4}=2\sqrt{3}~cm\\\underline{\sf c\acute alculo~do~ap\acute otema~da~pir\hat amide\!:}\\\sf A_l=6\cdot\dfrac{1}{2}\cdot a\cdot a_p\\\sf 30\diagdown\!\!\!\!\sqrt{3}=3\cdot2\diagdown\!\!\!\!\!\sqrt{3}\cdot a_p\\\sf a_p=\dfrac{30}{6}\\\sf a_p=5~cm\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf c\acute alculo~do~ap\acute otema~da~base\!:}\\\sf m=\dfrac{a\sqrt{3}}{2}\\\sf m=\dfrac{2\sqrt{3}\cdot\sqrt{3}}{2}\\\sf m=3~cm\\\underline{\sf c\acute alculo~da~altura\!:}\\\sf h^2+m^2=a_p^2\\\sf h^2+3^2=5^2\\\sf h^2+9=25\\\sf h^2=25-9\\\sf h^2=16\\\sf h=\sqrt{16}\\\sf h=4~cm\\\underline{\rm c\acute alculo~do~volume\!:}\\\sf V=\dfrac{1}{3}\cdot B\cdot h\\\sf V=\dfrac{1}{3}\cdot18\sqrt{3}\cdot 4\\\sf V=24~\sqrt{3}~cm^3\end{array}}$