Tetrahedral pyramid

Calculate the volume and surface of the regular tetrahedral pyramid if content area of the base is 20 cm2 and deviation angle of the side edges from the plane of the base is 60 degrees.

Result

V =  36.51 cm3
S =  72.915 cm2

Solution:

S1=20 cm2 Φ=60   S1=a2 a=S1=202 5 cm4.4721 cm  u=2 a=2 4.47212 10 cm6.3246 cm  h=u2 tanΦ=u2 tan60 =6.324555320342 tan60 =u2 1.732051=5.47723  V=13 S1 h=13 20 5.477236.514836.51 cm3S_{1}=20 \ \text{cm}^2 \ \\ Φ=60 \ ^\circ \ \\ \ \\ S_{1}=a^2 \ \\ a=\sqrt{ S_{1} }=\sqrt{ 20 } \doteq 2 \ \sqrt{ 5 } \ \text{cm} \doteq 4.4721 \ \text{cm} \ \\ \ \\ u=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 4.4721 \doteq 2 \ \sqrt{ 10 } \ \text{cm} \doteq 6.3246 \ \text{cm} \ \\ \ \\ h=\dfrac{ u }{ 2 } \cdot \ \tan Φ ^\circ =\dfrac{ u }{ 2 } \cdot \ \tan 60^\circ \ =\dfrac{ 6.32455532034 }{ 2 } \cdot \ \tan 60^\circ \ =\dfrac{ u }{ 2 } \cdot \ 1.732051=5.47723 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 20 \cdot \ 5.4772 \doteq 36.5148 \doteq 36.51 \ \text{cm}^3
b=h2+(a/2)2=5.47722+(4.4721/2)235 cm5.9161 cm  S2=a b/2=4.4721 5.9161/25 7 cm213.2288 cm2  S=4 S2+S1=4 13.2288+2072.91572.915 cm2b=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 5.4772^2 + (4.4721/2)^2 } \doteq \sqrt{ 35 } \ \text{cm} \doteq 5.9161 \ \text{cm} \ \\ \ \\ S_{2}=a \cdot \ b/2=4.4721 \cdot \ 5.9161/2 \doteq 5 \ \sqrt{ 7 } \ \text{cm}^2 \doteq 13.2288 \ \text{cm}^2 \ \\ \ \\ S=4 \cdot \ S_{2} + S_{1}=4 \cdot \ 13.2288 + 20 \doteq 72.915 \doteq 72.915 \ \text{cm}^2



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