Pentagonal Pyramid

Written by Jerry Ratzlaff on . Posted in Solid Geometry

 

pentagonal pyramid 2

pentagonal prism

  • 1 base
  • 10 edges
  • 5 side faces
  • 6 vertexs

Edge formula

\(a = 2  \sqrt { \frac {A_s ^2}        {         75h^2 - 25       \sqrt {5}   h^2 + A_s        \sqrt { {200 -} \sqrt {8000} }          }   \sqrt { {3 -} \sqrt {5} }    } \)

\(a = \left( 5 - \sqrt 5 \right) ^{1/4}   \sqrt { \sqrt {10} \frac {A_b} {5}   -   \sqrt {2} \frac {A_b} {5} } \)

\(a = \sqrt   {           24 \frac { V }   { 5h \left( \sqrt 2 + \sqrt {10} \right) }             }     { \left( 5- \sqrt 5 \right)^{1/4} } \)

Where:

\(a\) = edge

\(h\) = height

\(A_b\) = base area

\(A_s\) = surface area

\(V\) = volume

Height formula

\(h = 24V \frac { \sqrt { {5 -} \sqrt {5} } }         { 5a^2 \left( \sqrt {2} + \sqrt {10} \right) }   \)

\(h = \sqrt {\sqrt {\frac {1}{500} } +\frac {3}{50} }      \sqrt {   6 \left( \frac {A_s}{a} \right) ^2     \sqrt {20} \left( \frac {A_s}{a} \right) ^2     -A \sqrt { {50 -} \sqrt {5} }   }\)

Where:

\(h\) = height

\(a\) = edge

\(A_s\) = surface area

\(V\) = volume

Base Area formula

\(A_b =\frac{5}{4} tan \left( 54^\circ \right) a^2 \)

Where:

\(A_b\) = base area

\(a\) = edge

\(tan\) = tangent

Face Area formula

\(A_f =   \frac{a}{2} \sqrt { h^2 + \left( \frac { a\; tan \left( 54^\circ \right)} {2} \right) ^2 }\)

Where:

\(A_f\) = face area

\(a\) = edge

\(h\) = height

\(tan\) = tangent

Surface Area formula

\(A_s= \frac{5}{4} tan \left( 54^\circ \right) a^2 + 5 \frac {a}{2} \sqrt { h^2 + \left( \frac { a\; tan \left( 54^\circ \right)} {2} \right) ^2  } \)

Where:

\(A_s\) = surface area

\(a\) = edge

\(h\) = height

\(tan\) = tangent

Volume formula

\(V=\frac{5}{12} tan \left( 54^\circ \right) ha^2 \)

Where:

\(V\) = volume

\(a\) = edge

\(h\) = height

\(tan\) = tangent