Triangular pyramid

It is given perpendicular regular triangular pyramid: base side a = 5 cm, height v = 8 cm, volume V = 28.8 cm3. What is it content (surface area)?

Result

S =  71.79 cm2

Solution:

a=5 cm h=8 cm V=28.8 cm3  h1=a2−(a/2)2=52−(5/2)2≐4.3301 cm S1=a⋅ h12=5⋅ 4.33012≐10.8253 cm2  h22=h2+(h1/3)2 h2=h2+(h1/3)2=82+(4.3301/3)2≐8.1292 cm  S2=a⋅ h22=5⋅ 8.12922≐20.3229 cm2  S=S1+3⋅ S2=10.8253+3⋅ 20.3229≐71.7941=71.79 cm2a = 5 \ cm \ \\ h = 8 \ cm \ \\ V = 28.8 \ cm^3 \ \\ \ \\ h_{ 1 } = \sqrt{ a^2 - (a/2)^2 } = \sqrt{ 5^2 - (5/2)^2 } \doteq 4.3301 \ cm \ \\ S_{ 1 } = \dfrac{ a \cdot \ h_{ 1 } }{ 2 } = \dfrac{ 5 \cdot \ 4.3301 }{ 2 } \doteq 10.8253 \ cm^2 \ \\ \ \\ h_{ 2 }^2 = h^2 + (h_{ 1 }/3)^2 \ \\ h_{ 2 } = \sqrt{ h^2 + (h_{ 1 }/3)^2 } = \sqrt{ 8^2 + (4.3301/3)^2 } \doteq 8.1292 \ cm \ \\ \ \\ S_{ 2 } = \dfrac{ a \cdot \ h_{ 2 } }{ 2 } = \dfrac{ 5 \cdot \ 8.1292 }{ 2 } \doteq 20.3229 \ cm^2 \ \\ \ \\ S = S_{ 1 } + 3 \cdot \ S_{ 2 } = 10.8253 + 3 \cdot \ 20.3229 \doteq 71.7941 = 71.79 \ cm^2

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Pythagorean theorem is the base for the right triangle calculator.
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See also our trigonometric triangle calculator.

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