Question

calcule a area de uma piramide de base quadrada de 5 cm de base 9cm para arestas laterais​

Answer 1

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$\boxed{\begin{array}{l}\underline{\rm c\acute alculo~do~ap\acute otema~da~pir\hat amide\!:}\\\sf 9^2=a_p^2+\bigg(\dfrac{5}{2}\bigg)^2\\\sf 81=a_p^2+\dfrac{25}{4}\\\sf324=4a_p^2+25\\\sf 4a_p^2=324-25\\\sf 4a_p^2=299\\\sf a_p=\sqrt{\dfrac{299}{4}}\\\sf a_p=\dfrac{\sqrt{299}}{2}~cm\\\underline{\rm c\acute alculo~da~altura~da~pir\hat amide\!:}\\\sf h^2+\bigg(\dfrac{5}{2}\bigg)^2=a_p^2\\\sf h^2+\dfrac{25}{4}=\dfrac{299}{4}\\\sf h^2=\dfrac{299-25}{4}\\\sf h^2=\dfrac{274}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\sf h=\sqrt{\dfrac{274}{4}}=\dfrac{\sqrt{274}}{2}~cm\\\sf A_{l ateral}=\diagup\!\!\!4\cdot\dfrac{1}{\diagup\!\!\!2}\cdot5\cdot\dfrac{\sqrt{299}}{\diagup\!\!\!2}\\\sf A_{lateral}=5\sqrt{299}~cm^2\\\sf A_{base}=5^2=25~cm^2\\\sf A_{total}=A_{lateral}+A_{base}\\\sf A_{total}=5\sqrt{299}+25\\\sf A_{total}=5\cdot(\sqrt{299}+5)~cm^2\end{array}}$