Over Lesson 1–6

A. pentagon

B. heptagon

C. octagon

D. decagon

Name polygon A by its number of sides.

Over Lesson 1–6

A. pentagon

B. hexagon

C. heptagon

D. octagon

Name polygon B by its number of sides.

Over Lesson 1–6

A. 25 cm

B. 35 cm

C. 40 cm

D. 45 cm

Find the perimeter of polygon A.

Over Lesson 1–6

A. 40 in.

B. 42 in.

C. 45 in.

D. 85 in.

Find the perimeter of polygon B.

Over Lesson 1–6

A. polygon A: regularpolygon B: regular

B. polygon A: regularpolygon B: irregular

C. polygon A: irregularpolygon B: regular

D. polygon A: irregularpolygon B: irregular

Classify the polygons as regular or irregular.

Over Lesson 1–6

A. 18 meters

B. 10 meters

C. 11.25 meters

D. 15 meters

A regular hexagon has a perimeter of 90 meters. What is the length of one side of the hexagon?

Content Standards

G.GMD.3 Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

Mathematical Practices

2 Reason abstractly and quantitatively.

6 Attend to precision.

Students will be able

• Identify and name three-dimensional figures.

• Find surface area and volume.

• Polyhedron- a solid with all flat surfaces that enclose a single region of space

• Face- the flat surface which is a polygon

• Edges- the line segments were the faces intersect

• Vertex- the point where three or more edges intersect

Polyhedrons (polyhedra) are named by the shape of their bases.

Identify Solids

A. Determine whether the solid is a polyhedron. Then identify the solid. If it is a polyhedron, name the bases, faces, edges, and vertices.

Identify Solids

The solid is formed by polygonal faces, so it is a polyhedron. The bases are rectangles. This solid is a rectangular prism.

Answer: rectangular prism;Bases: rectangles EFHG, ABDC

Faces: rectangles FBDH, EACG, GCDH,EFBA, EFHG, ABDC

Vertices: A, B, C, D, E, F, G, H

Identify Solids

B. Determine whether the solid is a polyhedron. Then identify the solid. If it is a polyhedron, name the bases, faces, edges, and vertices.

Identify Solids

The solid is formed by polygonal faces, so it is a polyhedron. The bases are hexagons. This solid is a hexagonal prism.

Answer: hexagonal prism;Bases: hexagon EFGHIJ and hexagon KLMNOP

Faces: rectangles EFLK, FGML, GHNM, HNOI, IOPJ, JPKE; hexagons EFGHIJ

and KLMNOP

Vertices: E, F, G, H, I, J, K, L, M, N, O, P

Identify Solids

C. Determine whether the solid is a polyhedron. Then identify the solid. If it is a polyhedron, name the bases, faces, edges, and vertices.

Identify Solids

The solid has a curved surface, so it is not a polyhedron. The base is a circle and there is one vertex. So, it is a cone.

Answer: Base: circle TVertex: Wno faces or edges

A. triangular pyramid

B. pentagonal prism

C. rectangular prism

D. square pyramid

A. Identify the solid.

A. cone

B. cylinder

C. pyramid

D. polyhedron

B. Identify the solid.

A. triangular prism

B. triangular pyramid

C. rectangular pyramid

D. cone

C. Identify the solid.

•Regular polyhedron- a polyhedron if all of its faces are regular congruent polygons and all of the edges are congruent.•Platonic Solids- 5 regular polyhedrons …named because Plato used them extensively

•Surface Area- a two dimensional measurement of the surface of a solid figure. It is the sum of the areas of each face.•Volume- the measure of the amount of space enclosed by a solid figure.

Find Surface Area and Volume

Find the surface area and volume of the cone.

π π

Use a calculator.

.

Find Surface Area and Volume

Answer: The cone has a surface area of about 75.4 cm2 and a volume of about 37.7 cm3.

r = 3, h = 4

Volume of a cone

Simplify.

Use a calculator.

A. surface area = 288 ft2

volume = 336 ft3

B. surface area = 336 ft2

volume = 288 ft3

C. surface area = 26 ft2

volume = 60 ft3

D. surface area = 488 ft2

volume = 122 ft3

Find the surface area and volume of the triangular prism.

Surface Area and Volume

A. CONTAINERS Mike is creating a mailing tubewhich can be used to mail posters andarchitectural plans. The diameter of the base is

inches, and the height is feet. Find the

amount of cardboard Mike needs to make the tube.

The amount of material used to make the tube would be equivalent to the surface area of the cylinder.

Surface Area and Volume

Surface area of a cylinder

r = 1.875 in., h = 32 in.

Answer: Mike needs about 399.1 square inches ofcardboard to make the tube.

Use a calculator.399.1

Surface Area and Volume

B. CONTAINERS Mike is creating a mailing tubewhich can be used to mail posters andarchitectural plans. The diameter of the base is

inches, and the height is feet. Find the

volume of the tube.

Volume of a cylinder

r = 1.875 in., h = 32 in.

Use a calculator.353.4

Surface Area and Volume

Answer: The volume of the tube is about 353.4 cubic inches.

A. surface area = 2520 in2

B. surface area = 18 in2

C. surface area = 180 in2

D. surface area = 1144 in2

A. Jenny has some boxes for shipping merchandise. Each box is in the shape of a rectangular prism with a length of 18 inches, a width of 14 inches, and a height of 10 inches. Find the surface area of the box.

A. volume = 1144 in3

B. volume = 14 in3

C. volume = 2520 in3

D. volume = 3600 in3

B. Jenny has some boxes for shipping merchandise. Each box is in the shape of a rectangular prism with a length of 18 inches, a width of 14 inches, and a height of 10 inches. Find the volume of the box.