12.1 & 12.2 – Explore Solids & Surface Area of Prisms and Cones.

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12.1 & 12.2 – Explore Solids & Surface Area of Prisms and Cones

Polyhedron:

A solid that is bounded by polygons

Faces:

Polygon on the side of the shape

Ex: Hex ABCDFE

Quad EFKL

Edges:

Where two polygons meet to form a line

Ex: EF

FK

Vertex:

Where 3 polygons meet to form a point

Ex: E

K

Non-Polyhedron:

An edge that isn’t a polygon

Base: Polygon the solid is named after.

Lateral Faces:

Parallelograms or triangles on the sides of the solid

Prism:

Polyhedron with two parallel, congruent basesNamed after its base

Pyramid:

Polyhedron with one base and lateral facesNamed after its base.

Regular: All of the faces are congruent regular polygons

Convex: Any two points on its surface can be connected by a segment that lies entirely inside or on the solid

Concave: A side of the solid goes inward

Cross Section:

Intersection of a plane and a solid

Euler’s Theorem:

Faces + Vertices = Edges + 2

F + V = E + 2

Platonic Solids:

Regular Polyhedra, only 5. Named after how many faces they have

Regular Tetrahedron: 4 faces

Cube: 6 faces

Regular Octahedron: 8 faces

Regular Dodecahedron: 12 faces

Regular Icosahedron: 20 faces

Determine whether the solid is a polyhedron. If it is, name the polyhedron and state the number of faces, vertices, and edges.

Rectangular prism

Polyhedron: YES or NO

Faces: ___________

Vertices: _________

Edges: ___________

6

8

12F + V = E + 2

6 + 8 = 12 + 214 = 14

Determine whether the solid is a polyhedron. If it is, name the polyhedron and state the number of faces, vertices, and edges.

curved sides

Polyhedron: YES or NO

Faces: ___________

Vertices: _________

Edges: ___________

Determine whether the solid is a polyhedron. If it is, name the polyhedron and state the number of faces, vertices, and edges.

Pentagonal Pyramid

Polyhedron: YES or NO

Faces: ___________

Vertices: _________

Edges: ___________

6

6

10F + V = E + 2

6 + 6 = 10 + 212 = 12

Determine whether the solid is a polyhedron. If it is, name the polyhedron and state the number of faces, vertices, and edges.

Triangular prism

Polyhedron: YES or NO

Faces: ___________

Vertices: _________

Edges: ___________

5

6

9F + V = E + 2

5 + 6 = 9 + 211 = 11

Determine whether the solid is a polyhedron. If it is, name the polyhedron and state the number of faces, vertices, and edges.

curved side

Polyhedron: YES or NO

Faces: ___________

Vertices: _________

Edges: ___________

Use Euler’s Theorem to find the value of n.

F + V = E + 2

n + 8 = 12 + 2n + 8 = 14

n = 6

Use Euler’s Theorem to find the value of n.

F + V = E + 2

5 + 6 = n + 211 = n + 2

9 = n

Use Euler’s Theorem to find the value of n.

F + V = E + 2

8 + n = 18 + 28 + n = 20

n = 12

Sketch the polyhedron.

Cube

Sketch the polyhedron.

Rectangular prism

Sketch the polyhedron.

Pentagonal pyramid

Determine if the solid is convex or concave.

convex

Determine if the solid is convex or concave.

concave

Determine if the solid is convex or concave.

convex

Describe the cross section formed by the intersection of the plane and the solid.

pentagon

Describe the cross section formed by the intersection of the plane and the solid.

circle

Describe the cross section formed by the intersection of the plane and the solid.

triangle

Cylinder: Prism with circular bases

Surface area: Area of each face of solid

Lateral area: Area of each lateral face

Right Prism: Each lateral edge is perpendicular to both bases

Oblique Prism: Each lateral edge is NOT perpendicular to both bases

Net: Two-dimensional representation of a solid

Surface Area of a Right Prism:

SA = 2B + PH

B = area of one base

P = Perimeter of one base

H = Height of the prism

H

Surface Area of a Right Cylinder:

22 2SA r rHπ π= +

H

SA = 2B + PH

1. Name the solid that can be formed by the net.

Cylinder

1. Name the solid that can be formed by the net.

Triangular prism

1. Name the solid that can be formed by the net.

rectangular prism

Cube?

2. Find the surface area of the right solid.

SA = 2B + PH

SA = 2(30) + (22)(7)

B = bhB = (5)(6)

B = 30

P = 5 + 6 + 5 + 6P = 22

SA = 60 + 154

SA = 214 m2

2. Find the surface area of the right solid.

SA = 2B + PH

SA = 2(30) + (30)(10)

P = 5 + 12 + 13P = 30

SA = 60 + 300

SA = 360 cm2

1

2B bh=

1(12)(5)

2B =

30B =

c2 = a2 + b2

c2 = (5)2 + (12)2

c2 = 25 + 144

c2 = 169

c = 13

2. Find the surface area of the right solid.

22 2SA r rHπ π= +22 (2) 2 (2)(6)SA π π= +

2 (4) 2 (12)SA π π= +

8 24SA π π= +

32SA π= cm2

2. Find the surface area of the right solid.

22 2SA r rHπ π= +22 (28) 2 (28)(144)SA π π= +

2 (784) 2 (4032)SA π π= +

1568 8064SA π π= +

9632SA π= in2

144in

3. Solve for x, given the surface area.

SA = 2B + PH

142 = 2(5x) + (2x + 10)(7)

B = bhB = 5x

P = 5 + x + 5 + xP = 2x + 10

142 = 10x + 14x + 70

142 = 24x + 70

72 = 24x

3ft = x

3. Solve for x, given the surface area.

22 2SA r rHπ π= +2326.73 2 (4) 2 (4)xπ π= +

326.73 2 (16) 8 xπ π= +326.73 32 8 xπ π= +

226.199 ≈8πx

9cm ≈x