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# How many vertices, edges, and faces of the polyhedron are there List them. - PowerPoint PPT Presentation

There are 10 vertices:. A , B , C , D , E , F , G , H , I , and J . There are 15 edges:. AF, BG, CH, DI, EJ, AB, BC, CD, DE, EA, FG, GH, HI, IJ, and JF. There are 7 faces:. pentagons: ABCDE and FGHIJ , and quadrilaterals: ABGF , BCHG , CDIH , DEJI , and EAFJ.

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Presentation Transcript

A, B, C, D, E, F, G, H, I, and J.

There are 15 edges:

AF, BG, CH, DI, EJ, AB, BC, CD,

DE, EA, FG, GH, HI, IJ, and JF.

There are 7 faces:

pentagons: ABCDE and FGHIJ, and

quadrilaterals: ABGF, BCHG, CDIH, DEJI, and EAFJ

Space Figures and Cross Sections

LESSON 11-1

How many vertices, edges, and faces of the

polyhedron are there? List them.

Quick Check

LESSON 11-1

Use Euler’s Formula to find the number of edges of a polyhedron with 6 faces and 8 vertices.

F+V= E+ 2 Euler’s Formula

6 + 8 = E+ 2 Substitute the number of faces and vertices.

12 = ESimplify.

A solid with 6 faces and 8 vertices has 12 edges.

Quick Check

Space Figures and Cross Sections

LESSON 11-1

Use the pentagonal prism from Example 1 to verify

Euler’s Formula. Then draw a net for the figure and verify

Euler’s Formula for the two-dimensional figure.

Use the faces F = 7, vertices V = 10, and edges E = 15.

F+V= E+ 2 Euler’s Formula

7 + 10 = 15 + 2 Substitute the number of faces and vertices.

Count the regions: F = 7

Count the vertices: V = 18

Count the segments: E = 24

F + V = E + 1 Euler’s Formula in two dimensions

7 + 18 = 24 + 1 Substitute.

Quick Check

LESSON 11-1

Describe this cross section.

The plane is parallel to the triangular base of the figure, so the cross section is also a triangle.

Quick Check

LESSON 11-1