Right Pentagonal Pyramid

Written by Jerry Ratzlaff on . Posted in Solid Geometry

  • pentagonal pyramid 2right pentagonal pyramid 3Right pentagonal pyramid (a three-dimensional figure) has pentagon base and the apex alligned above the center of the base.
  • 1 base
  • 10 edges
  • 5 side faces
  • 6 vertexs

Base Area of a Right Pentagonal Pyramid formula

\(\large{ A_b = \frac{5}{4}\; tan\;54° \;a^2 }\)

Where:

\(\large{ A_b }\) = base area

\(\large{ a }\) = edge

\(\large{ tan }\) = tangent

Edge of a Right Pentagonal Pyramid formula

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

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

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

Where:

\(\large{ a }\) = edge

\(\large{ h }\) = height

\(\large{ A_b }\) = base area

\(\large{ A_s }\) = surface area

\(\large{ V }\) = volume

Height of a Right Pentagonal Pyramid formula

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

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

Where:

\(\large{ h }\) = height

\(\large{ a }\) = edge

\(\large{ A_s }\) = surface area

\(\large{ V }\) = volume

Face Area of a Right Pentagonal Pyramid formula

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

Where:

\(\large{ A_f }\) = face area

\(\large{ a }\) = edge

\(\large{ h }\) = height

\(\large{ tan }\) = tangent

Surface Area of a Right Pentagonal Pyramid formula

\(\large{ A_s = \frac{5}{2}\; \left( r \; a \right) + \frac{5}{2}\; \left( a \; h_s \right)  }\)

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

Where:

\(\large{ A_s }\) = surface area

\(\large{ a }\) = edge

\(\large{ h_s }\) = height side

\(\large{ r }\) = radius

Volume of a Right Pentagonal Pyramid formula

\(\large{ V= \frac{5}{6}\; r\;a\;h }\)

\(\large{ V= \frac{5}{12}\; tan\;54° \;a^2\;h }\)

Where:

\(\large{ V }\) = volume

\(\large{ a }\) = edge

\(\large{ h }\) = height

\(\large{ tan }\) = tangent

 

Tags: Equations for Volume