medida a. O volume desta pirâmide é?
distância (x), do centro da base à aresta lateral:
2x=a√2............2x=diagonal do quadrado(da base)
x=a√2/2 >
a²=x²+h²
h²=a²-x²
h²=a² -(a√2/2)²
h²=a² -2a²/4
h²=2a²/4
h=√2a²/4
h=a√2 /2 >>>
v=sb .h/3
v=a² .(a√2 /2) /3
v=a³√2 /6 cm³ >>>>>>>>>>>>>>>
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distância (x), do centro da base à aresta lateral:
2x=a√2............2x=diagonal do quadrado(da base)
x=a√2/2 >
a²=x²+h²
h²=a²-x²
h²=a² -(a√2/2)²
h²=a² -2a²/4
h²=2a²/4
h=√2a²/4
h=a√2 /2 >>>
v=sb .h/3
v=a² .(a√2 /2) /3
v=a³√2 /6 cm³ >>>>>>>>>>>>>>>