Tetrahedral pyramid

Calculate the volume and surface area of a regular tetrahedral pyramid, its height is $b cm and the length of the edges of the base is 6 cm.

Result

S =  196.1 cm2
V =  156 cm3

Solution:

a=6 cm h=13 cm s=h2+(a/2)2=132+(6/2)2≐178 cm≐13.3417 cm S1=a⋅ s/2=6⋅ 13.3417/2≐3 178 cm2≐40.025 cm2  S=a2+4⋅ S1=62+4⋅ 40.025≐196.1≐196.1 cm2a=6 \ \text{cm} \ \\ h=13 \ \text{cm} \ \\ s=\sqrt{ h^2+(a/2)^2 }=\sqrt{ 13^2+(6/2)^2 } \doteq \sqrt{ 178 } \ \text{cm} \doteq 13.3417 \ \text{cm} \ \\ S_{1}=a \cdot \ s/2=6 \cdot \ 13.3417/2 \doteq 3 \ \sqrt{ 178 } \ \text{cm}^2 \doteq 40.025 \ \text{cm}^2 \ \\ \ \\ S=a^2 + 4 \cdot \ S_{1}=6^2 + 4 \cdot \ 40.025 \doteq 196.1 \doteq 196.1 \ \text{cm}^2
V=13⋅ a2⋅ h=13⋅ 62⋅ 13=156 cm3V=\dfrac{ 1 }{ 3 } \cdot \ a^2 \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 6^2 \cdot \ 13=156 \ \text{cm}^3

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