Pyramid 4sides

Calculate the volume and the surface of a regular quadrangular pyramid when the edge of the base is 4 cm long and the height of the pyramid is 7 cm.

Result

V =  37.333 cm3
S =  74.241 cm2

Solution:

a=4 cm h=7 cm  S1=a2=42=16 cm2  V=S1⋅ h/3=16⋅ 7/3≐1123≐37.3333≐37.333 cm3a=4 \ \text{cm} \ \\ h=7 \ \text{cm} \ \\ \ \\ S_{1}=a^2=4^2=16 \ \text{cm}^2 \ \\ \ \\ V=S_{1} \cdot \ h/3=16 \cdot \ 7/3 \doteq \dfrac{ 112 }{ 3 } \doteq 37.3333 \doteq 37.333 \ \text{cm}^3
h2=h2+(a/2)2=72+(4/2)2≐53 cm≐7.2801 cm S2=a⋅ h2/2=4⋅ 7.2801/2≐2 53 cm2≐14.5602 cm2  S=S1+4⋅ S2=16+4⋅ 14.5602≐74.2409≐74.241 cm2h_{2}=\sqrt{ h^2+(a/2)^2 }=\sqrt{ 7^2+(4/2)^2 } \doteq \sqrt{ 53 } \ \text{cm} \doteq 7.2801 \ \text{cm} \ \\ S_{2}=a \cdot \ h_{2} / 2=4 \cdot \ 7.2801 / 2 \doteq 2 \ \sqrt{ 53 } \ \text{cm}^2 \doteq 14.5602 \ \text{cm}^2 \ \\ \ \\ S=S_{1} + 4 \cdot \ S_{2}=16 + 4 \cdot \ 14.5602 \doteq 74.2409 \doteq 74.241 \ \text{cm}^2

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