Wall height

Calculate the surface and volume of a regular quadrangular pyramid if side a = 6 cm and wall height v = 0.8dm.

Correct result:

S =  132 cm2
V =  88.994 cm3

Solution:

a=6 v=0.8⋅ 10=8 S1=a2=62=36 S2=a⋅ v/2=6⋅ 8/2=24 S=S1+4⋅ S2=36+4⋅ 24=132 cm2a=6 \ \\ v=0.8 \cdot \ 10=8 \ \\ S_{1}=a^2=6^2=36 \ \\ S_{2}=a \cdot \ v/2=6 \cdot \ 8/2=24 \ \\ S=S_{1} + 4 \cdot \ S_{2}=36 + 4 \cdot \ 24=132 \ \text{cm}^2
h=v2−(a/2)2=82−(6/2)2≐55≐7.4162 V=S1⋅ h/3=36⋅ 7.4162/3=12 55=88.994 cm3h=\sqrt{ v^2-(a/2)^2 }=\sqrt{ 8^2-(6/2)^2 } \doteq \sqrt{ 55 } \doteq 7.4162 \ \\ V=S_{1} \cdot \ h/3=36 \cdot \ 7.4162/3=12 \ \sqrt{ 55 }=88.994 \ \text{cm}^3

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