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).

Result

V =  83.467 cm3
S =  87.87 cm2

Solution:

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 5.926683.4668=83.467 cm3a = 6.5 \ cm \ \\ s = 7.5 \ cm \ \\ \ \\ S_{ 1 } = a^2 = 6.5^2 = \dfrac{ 169 }{ 4 } = 42.25 \ cm^2 \ \\ \ \\ u_{ 1 } = \sqrt{ 2 } \cdot \ a = \sqrt{ 2 } \cdot \ 6.5 \doteq 9.1924 \ cm \ \\ \ \\ h = \sqrt{ s^2 - (u_{ 1 }/2)^2 } = \sqrt{ 7.5^2 - (9.1924/2)^2 } \doteq 5.9266 \ cm \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ S_{ 1 } \cdot \ h = \dfrac{ 1 }{ 3 } \cdot \ 42.25 \cdot \ 5.9266 \doteq 83.4668 = 83.467 \ 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 21.967687.8703=87.87 cm2h_{ 2 } = \sqrt{ h^2 + (a/2)^2 } = \sqrt{ 5.9266^2 + (6.5/2)^2 } \doteq 6.7593 \ cm \ \\ \ \\ S_{ 2 } = \dfrac{ 1 }{ 2 } \cdot \ a \cdot \ h_{ 2 } = \dfrac{ 1 }{ 2 } \cdot \ 6.5 \cdot \ 6.7593 \doteq 21.9676 \ cm^2 \ \\ \ \\ S = 4 \cdot \ S_{ 2 } = 4 \cdot \ 21.9676 \doteq 87.8703 = 87.87 \ cm^2



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Tip: Our volume units converter will help you with the conversion of volume units. Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.

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