Question

Numa piramide triangular regular a aresta da base mede 6m e a aresta lateral 8 m. Calcular: a) o apótema da pirámide; b) o apótema da base; c) a altura da piramide: d) área total da piramide

Answer 1

Explicação passo-a-passo:

a)

$\sf (a_L)^2=\Big(\dfrac{L}{2}\Big)^2+(a_p)^2$

$\sf 8^2=\Big(\dfrac{6}{2}\Big)^2+(a_p)^2$

$\sf 8^2=3^2+(a_p)^2$

$\sf 64=9+(a_p)^2$

$\sf (a_p)^2=64-9$

$\sf (a_p)^2=55$

$\sf \red{a_p=\sqrt{55}~m}$

b)

$\sf a_{p_{b}}=\dfrac{L\sqrt{3}}{2}\cdot\dfrac{1}{3}$

$\sf a_{p_{b}}=\dfrac{6\sqrt{3}}{2}\cdot\dfrac{1}{3}$

$\sf a_{p_{b}}=\dfrac{6\sqrt{3}}{6}$

$\sf \red{a_{p_{b}}=\sqrt{3}~m}$

c)

$\sf (a_p)^2=(a_{p_{b}})^2+h^2$

$\sf (\sqrt{55})^2=(\sqrt{3})^2+h^2$

$\sf 55=3+h^2$

$\sf h^2=55-3$

$\sf h^2=52$

$\sf h=\sqrt{52}$

$\sf \red{h=2\sqrt{13}~m}$

d)

$\sf A_{T}=A_{b}+A_{L}$

$\sf A_{T}=\dfrac{6^2\sqrt{3}}{4}+3\cdot\dfrac{6\cdot\sqrt{55}}{2}$

$\sf A_{T}=\dfrac{36\sqrt{3}}{4}+3\cdot3\sqrt{55}$

$\sf A_{T}=9\sqrt{3}+9\sqrt{55}$

$\sf \red{A_{T}=9\cdot(\sqrt{3}+\sqrt{55})~m^2}$