Question

A base de uma pirâmide regular é um triângulo equilátero de perímetro igual a 18 cm e a altura da pirâmide mede 9 cm. Qual a área total desta pirâmide?​

Answer 1

$\large\boxed{\begin{array}{l}\rm 3\ell=18\\\rm \ell=\dfrac{18}{3}\\\\\rm \ell=6\,cm\\\rm B=\dfrac{\ell^2\sqrt{3}}{4}=\dfrac{6^2\sqrt{3}}{4}=9\sqrt{3}\,cm^2\\\\\rm m=\dfrac{\ell\sqrt{3}}{6}=\dfrac{\backslash\!\!\!6\sqrt{3}}{\backslash\!\!\!6}=\sqrt{3}\\\rm a_p^2=9^2+\sqrt{3^2}\\\\\rm a_p^2=81+3\\\rm a_p^2=84\\\rm a_p=\sqrt{84}\\\rm a_p=\sqrt{4\cdot21}\\\rm a_p=2\sqrt{21}\,cm\end{array}}$

$\large\boxed{\begin{array}{l}\rm A_l=3\cdot\dfrac{1}{\backslash\!\!\!2}\cdot6\cdot\backslash\!\!\!2\sqrt{21}\\\rm A_l=18\sqrt{21}\,cm^2\\\rm A_t=B+A_l\\\rm A_t=9\sqrt{3}+18\sqrt{21}\\\rm A_t=9\sqrt{3}\cdot(1+2\sqrt{7})\,cm^2\end{array}}$