Question

Uma pirâmide quadrangular regular tem altura igual a √7 cm e apótema da base igual a 3√2 cm. Calcular:
a) a aresta da base
b) a apótema da pirâmide
c) a aresta lateral
d) a área da base
e) a área lateral
f) a área total
g) o volume

Answer 1

$\huge\boxed{\boxed{\boxed{\mathfrak{Legenda:}}}}$

$\mathsf{a}$→aresta da base

$\mathsf{a_{p}}$ →apótema da pirâmide

$\mathsf{\ell}$ →aresta lateral

$\mathsf{B}$→área da base

$\mathsf{A_{L}}$→área lateral

$\mathsf{A_{t}}$→área total

$\mathsf{V}$→volume da pirâmide

$\mathsf{m}$ →apótema da base

$\mathsf{h}$→altura

Solução :

a)

$\mathsf{m=\dfrac{a}{2}}\\\mathsf{3\sqrt{2}=\dfrac{a}{2}\to~a=2.3\sqrt{2}}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{a=6\sqrt{2}~cm}}}}}$

b)

$\mathsf{a_{p}^2=h^2+m^2}\\\mathsf{a_{p}^2=(\sqrt{7})^2+(3\sqrt{2})^2}\\\mathsf{a_{p}^2=7+18}\\\mathsf{a_{p}^2=25}$

$\mathsf{a_{p}=\sqrt{25}}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{a_{p}=5~cm}}}}}$

c)

$\mathsf{\ell^2=a_{p}^2+(\dfrac{a}{2})^2}\\\mathsf{\ell^2=25+(3\sqrt{2})^2}\\\mathsf{\ell^2=25+18}\\\mathsf{\ell^2=43}$

$\large\boxed{\boxed{\boxed{\boxed{\mathsf{\ell=\sqrt{43}~cm}}}}}$

d)

$\mathsf{B=a^2}\\\mathsf{B=(6\sqrt{2}}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{B=72~cm^2}}}}}$

e)

$\mathsf{A_{L}=4.\dfrac{1}{2}.a.a_{p}}\\\mathsf{A_{L}=2.6\sqrt{2}.5}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{A_{L}=60\sqrt{2}~cm^2}}}}}$

f)

$\mathsf{A_{t}=A_{L}+B}\\\mathsf{A_{t}=60\sqrt{2}+72}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{A_{t}=12(6+5\sqrt{2})~cm^2}}}}}$

g)

$\mathsf{V=\dfrac{1}{3}.B.h}\\\mathsf{V=\dfrac{1}{3}.72.\sqrt{7}}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{V=24\sqrt{7}~cm^3}}}}}$