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:

$a = 6 \ cm \ \\ h = 13 \ cm \ \\ s = \sqrt{ h^2+(a/2)^2 } = \sqrt{ 13^2+(6/2)^2 } = \sqrt{ 178 } \ cm \doteq 13.3417 \ cm \ \\ S_{ 1 } = a \cdot \ s/2 = 6 \cdot \ 13.3417/2 = 3 \ \sqrt{ 178 } \ cm^2 \doteq 40.025 \ cm^2 \ \\ \ \\ S = a^2 + 4 \cdot \ S_{ 1 } = 6^2 + 4 \cdot \ 40.025 \doteq 196.1 = 196.1 \ cm^2$
$V = \dfrac{ 1 }{ 3 } \cdot \ a^2 \cdot \ h = \dfrac{ 1 }{ 3 } \cdot \ 6^2 \cdot \ 13 = 156 = 156 \ cm^3$

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Pythagorean theorem is the base for the right triangle calculator. Do you want to convert area units? Do you know the volume and unit volume, and want to convert volume units? See also our trigonometric triangle calculator.

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