Question

em uma pirâmide regular triangular, cada aresta lateral mede 13 cm e cada aresta da base mede 10 cm. Calcular:
a) a medida m do apótema da pirâmide
b) a medida r do apótema da base da pirâmide
c) a medida H da altura da pirâmide​

Answer 1

Explicação passo-a-passo:

a)

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

$\sf 13^2=\Big(\dfrac{10}{2}\Big)+m^2$

$\sf 13^2=5^2+m^2$

$\sf 169=25+m^2$

$\sf m^2=169-25$

$\sf m^2=144$

$\sf m=\sqrt{144}$

$\sf \red{m=12~cm}$

b)

$\sf r=\dfrac{L\sqrt{3}}{2}\cdot\dfrac{1}{3}$

$\sf r=\dfrac{10\sqrt{3}}{2}\cdot\dfrac{1}{3}$

$\sf r=5\sqrt{3}\cdot\dfrac{1}{3}$

$\sf \red{r=\dfrac{5\sqrt{3}}{3}~cm}$

c)

$\sf m^2=r^2+h^2$

$\sf 12^2=\Big(\dfrac{5\sqrt{3}}{3}\Big)^2+h^2$

$\sf 144=\dfrac{75}{9}+h^2$

$\sf 144=\dfrac{75}{9}+h^2$

$\sf h^2=144-\dfrac{75}{9}$

$\sf h^2=\dfrac{1296-75}{9}$

$\sf h^2=\dfrac{1221}{9}$

$\sf h=\sqrt{\dfrac{1221}{9}}$

$\sf \red{h=\dfrac{\sqrt{1221}}{3}~cm}$