Tetrahedral pyramid

A regular tetrahedral pyramid is given. Base edge length a = 6.5 cm, side edge s = 7.5 cm. Calculate the volume and the area of its face (side area).

Correct result:

V =  83.467 cm3
S =  87.87 cm2


a=6.5 cm s=7.5 cm  S1=a2=6.52=1694=42.25 cm2  u1=2 a=2 6.59.1924 cm  h=s2(u1/2)2=7.52(9.1924/2)25.9266 cm  V=13 S1 h=13 42.25 5.9266=83.467 cm3a=6.5 \ \text{cm} \ \\ s=7.5 \ \text{cm} \ \\ \ \\ S_{1}=a^2=6.5^2=\dfrac{ 169 }{ 4 }=42.25 \ \text{cm}^2 \ \\ \ \\ u_{1}=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 6.5 \doteq 9.1924 \ \text{cm} \ \\ \ \\ h=\sqrt{ s^2 - (u_{1}/2)^2 }=\sqrt{ 7.5^2 - (9.1924/2)^2 } \doteq 5.9266 \ \text{cm} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 42.25 \cdot \ 5.9266=83.467 \ \text{cm}^3
h2=h2+(a/2)2=5.92662+(6.5/2)26.7593 cm  S2=12 a h2=12 6.5 6.759321.9676 cm2  S=4 S2=4 21.9676=87.87 cm2h_{2}=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 5.9266^2 + (6.5/2)^2 } \doteq 6.7593 \ \text{cm} \ \\ \ \\ S_{2}=\dfrac{ 1 }{ 2 } \cdot \ a \cdot \ h_{2}=\dfrac{ 1 }{ 2 } \cdot \ 6.5 \cdot \ 6.7593 \doteq 21.9676 \ \text{cm}^2 \ \\ \ \\ S=4 \cdot \ S_{2}=4 \cdot \ 21.9676=87.87 \ \text{cm}^2

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