Triangular pyramid

Determine the volume and surface area of a regular triangular pyramid having a base edge a=20 cm and a lateral edge b = 35 cm

Result

V =  1848.423 cm3
S =  1179.436 cm2

Solution:

a=20 cm b=35 cm  a2=a/2=20/2=10 cm h1=b2−a22=352−102≐15 5 cm≐33.541 cm h=h12−a22=33.5412−102≐5 41 cm≐32.0156 cm  S1=a⋅ h1/2=20⋅ 33.541/2≐150 5 cm2≐335.4102 cm2 S2=34⋅ a2=34⋅ 202≐100 3 cm2≐173.2051 cm2  V=13⋅ S2⋅ h=13⋅ 173.2051⋅ 32.0156≐1848.4228≐1848.423 cm3a=20 \ \text{cm} \ \\ b=35 \ \text{cm} \ \\ \ \\ a_{2}=a/2=20/2=10 \ \text{cm} \ \\ h_{1}=\sqrt{ b^2-a_{2}^2 }=\sqrt{ 35^2-10^2 } \doteq 15 \ \sqrt{ 5 } \ \text{cm} \doteq 33.541 \ \text{cm} \ \\ h=\sqrt{ h_{1}^2 - a_{2}^2 }=\sqrt{ 33.541^2 - 10^2 } \doteq 5 \ \sqrt{ 41 } \ \text{cm} \doteq 32.0156 \ \text{cm} \ \\ \ \\ S_{1}=a \cdot \ h_{1} / 2=20 \cdot \ 33.541 / 2 \doteq 150 \ \sqrt{ 5 } \ \text{cm}^2 \doteq 335.4102 \ \text{cm}^2 \ \\ S_{2}=\dfrac{ \sqrt{ 3 } }{ 4 } \cdot \ a^2=\dfrac{ \sqrt{ 3 } }{ 4 } \cdot \ 20^2 \doteq 100 \ \sqrt{ 3 } \ \text{cm}^2 \doteq 173.2051 \ \text{cm}^2 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{2} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 173.2051 \cdot \ 32.0156 \doteq 1848.4228 \doteq 1848.423 \ \text{cm}^3
S=3⋅ S1+S2=3⋅ 335.4102+173.2051≐1179.4357≐1179.436 cm2S=3 \cdot \ S_{1} + S_{2}=3 \cdot \ 335.4102 + 173.2051 \doteq 1179.4357 \doteq 1179.436 \ \text{cm}^2

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

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