Hexagonal pyramid

Calculate the volume and the surface of a regular hexagonal pyramid with a base edge length of 3 cm and a height of 5 cm.


V =  38.971 cm3
S =  74.095 cm2


a=3 cm h=5 cm S1=3 3/2 a2=3 3/2 3223.3827 cm2 V=S1 h/3=23.3827 5/338.971138.971 cm3a=3 \ \text{cm} \ \\ h=5 \ \text{cm} \ \\ S_{1}=3 \cdot \ \sqrt{ 3 }/2 \cdot \ a^2=3 \cdot \ \sqrt{ 3 }/2 \cdot \ 3^2 \doteq 23.3827 \ \text{cm}^2 \ \\ V=S_{1} \cdot \ h/3=23.3827 \cdot \ 5/3 \doteq 38.9711 \doteq 38.971 \ \text{cm}^3
a2=3/2 a=3/2 32.5981 cm h2=a22+h2=2.59812+525.6347 cm S2=a h2/2=3 5.6347/28.4521 cm2 S=S1+6 S2=23.3827+6 8.452174.095174.095 cm2a_{2}=\sqrt{ 3 }/2 \cdot \ a=\sqrt{ 3 }/2 \cdot \ 3 \doteq 2.5981 \ \text{cm} \ \\ h_{2}=\sqrt{ a_{2}^2+h^2 }=\sqrt{ 2.5981^2+5^2 } \doteq 5.6347 \ \text{cm} \ \\ S_{2}=a \cdot \ h_{2}/2=3 \cdot \ 5.6347/2 \doteq 8.4521 \ \text{cm}^2 \ \\ S=S_{1}+6 \cdot \ S_{2}=23.3827+6 \cdot \ 8.4521 \doteq 74.0951 \doteq 74.095 \ \text{cm}^2

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