Right Pentagonal Prism

Written by Jerry Ratzlaff on . Posted in Solid Geometry

  • pentagonal prism 2right pentagonal prism 1Right pentagonal prism (a three-dimensional figure) has pentagonal bases and each face is a regular polygon with equal sides and equal angles.
  • 20 base diagonals
  • 10 face diagonals
  • 20 space diagonals
  • 2 bases
  • 15 edges
  • 5 side faces
  • 10 vertexs

Base Area of a Right Pentagonal Prism formula

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

Where:

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

\(\large{ a }\) = edge

Edge of a Right Pentagonal Prism formula

\(\large{ a = 25^{3/4}   {\frac   {\sqrt {A_b} }     { 5\; \left( \sqrt {20}\; +5 \right) ^1/4 }   }  }\)

\(\large{ a = \frac   { - \left ( 5\; \sqrt {5} h  - \sqrt { 125\;h^2 + 10\; A_s\; \sqrt { \sqrt {500}\; +25 }   }   \right) }     { 5\; \sqrt { \sqrt { 20 }\; +5 }  }  }\)

\(\large{ a = \frac {A_l}   {5\;h} }\)

Where:

\(\large{ a }\) = edge

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

\(\large{ h }\) = height

\(\large{ A_l }\) = lateral surface area

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

Height of a Right Pentagonal Prism formula

\(\large{ h = \frac {A_s} {5\;a} -   \frac {1}{10}\;a\; \sqrt { \sqrt {500} +25 } }\)

Where:

\(\large{ h }\) = height

\(\large{ a }\) = edge

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

Lateral Surface Area of a Right Pentagonal Prism formula

\(\large{ A_l = 5\;a\;h }\)

Where:

\(\large{ A_l }\) = lateral surface area

\(\large{ a }\) = edge

\(\large{ h }\) = height

Surface Area of a Right Pentagonal Prism formula

\(\large{ A_s = 5\;a\;h + \frac {1}{2} \;  \sqrt { 5\; \left ( 5+2 \;\sqrt {5} \right) }\; a^2   }\)

Where:

\(\large{  A_s }\) = surface area

\(\large{ a }\) = edge

\(\large{ h }\) = height

Volume of a Right Pentagonal Prism formula

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

Where:

\(\large{ V }\) = volume

\(\large{ a }\) = edge

\(\large{ h }\) = height

 

Tags: Equations for Volume