Show that if there is an Archimedean solid that has faces that are regular pentagons and equilateral triangles, and where four faces meet at each vertex, then there must be eight more triangles than pentagons.
Show that if there is an Archimedean solid that has faces that are regular pentagons and equilateral triangles, and where four faces meet at each vertex, then there must be eight more triangles than pentagons.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.1: Rectangular Coordinate Systems
Problem 33E
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![Show that if there is an Archimedean solid that has faces that are regular pentagons
and equilateral triangles, and where four faces meet at each vertex, then there must
be eight more triangles than pentagons.
1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b3f819d-a822-4927-9810-35cb223e228d%2F576221a1-57c6-4165-9dde-e15c7db89990%2F99slnjq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show that if there is an Archimedean solid that has faces that are regular pentagons
and equilateral triangles, and where four faces meet at each vertex, then there must
be eight more triangles than pentagons.
1.
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