Chapter 11

Section 11.1 – Space Figures and Cross Sections

Objectives:To recognize polyhedra and their partsTo visualize cross sections of space figures

Polyhedron a three-dimensional figure whose surfaces are polygons

Face name for a side of a polyhedron

Edge a segment that is formed by the intersection of two faces

Vertex a point where three or more edges intersect

Faces

Edge

Vertex

Ex: ◦ Identify the vertices/edges/faces of the figure

A

BC

D

E

F G

H

Euler’s Formula◦ The numbers of faces (F), vertices (V), and edges

(E) of a polyhedron are related by the formula:

F + V = E + 2

Ex: Using Euler’s Formula

◦ Use Euler’s Formula to find the number of edges on a polyhedron with eight triangular faces.

Cross-Section the intersection of a solid and a plane. You can think of it as a very thin slice of the solid.

Examples of Cross Sections◦ One slice of bread in a loaf◦ CAT Scans and MRI’s

Homework #25Due Tuesday (April 09)Page 601◦# 1 – 19 all

Objectives:

To find the surface area of a prismTo find the surface area of a cylinder

Section 11.2 – Surface Areas of Prisms and Cylinders

Prism a polyhedron with exactly two congruent, parallel faces called bases. The other faces of the prism are called lateral faces.

Altitude of a prism is a perpendicular segment that joins the planes of the bases.

Height of the prism is the length of an altitude.

Lateral Area of a prism is the sum of the areas of the lateral faces.

Surface Area the sum of the lateral area of the area of the two bases.

Theorem 11.1 – Lateral and Surface Areas of a Prism

◦ The lateral area of a right prism is the product of the perimeter of the base and the height.

L.A. = p · h

◦ The surface area of a right prism is the sum of the lateral area and the areas of the two bases.

S.A. = L.A. + 2B

Cylinder has two congruent parallel bases, just like a prism. However, the bases of a cylinder are circles.

Altitude of a cylinder is a perpendicular segment that joins the planes of the bases

Height of a cylinder is the length of an altitude

Lateral Area visualize “unrolling” the curved surface of the cylinder. Imagine taking the label off of a water bottle. The area of the resulting rectangle is the lateral area.

Surface Area of a cylinder is the sum of the lateral area and the areas of the two circular bases.

Theorem 11.2 – Lateral and Surface Areas of a Cylinder

◦ The lateral area of a right cylinder is the product of the circumference of the base and the height of the cylinder.

L.A. = 2Πrh -or- L.A. =Πdh

◦ The surface area of a right cylinder is the sum of the lateral area and the areas of the two bases.

S.A. = L.A. + 2B -or- S.A. = 2Πrh + 2Π

Homework # 26Due Wednesday (April 10)

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Objectives:

To find the surface area of a pyramidTo find the surface area of a cone

Section 11.3 – Surface Areas of Pyramids and Cones

Pyramid a polyhedron in which one face (the base) can be any polygon and the other faces (lateral faces) are triangles that meet at a common vertex (vertex of the pyramid).

Altitude of a pyramid is the perpendicular segment from the vertex to the plane of the base.

Height length of the altitude

Regular Pyramid a pyramid whose base is a regular polygon and whose lateral faces are congruent isosceles triangles.

Slant height (l) the length of the altitude of a lateral face of the pyramid.

Lateral Area of a pyramid is the sum of the areas of the congruent lateral faces.

Surface Area of a pyramid is the sum of the lateral area and the area of its base

Theorem 11.3 – Lateral and Surface Areas of a Regular Pyramid

◦ The lateral area of a regular pyramid is half the product of the perimeter of the base and the slant height.

L.A. = p · l

◦ The surface area of a regular pyramid is the sum of the lateral area and the area of the base.

S.A. = L.A. + B

Cone has a pointed top like a pyramid, but its base is a circle

Altitude a perpendicular segment from the vertex of the cone to the center of its base

Height the length of the altitude

Slant Height the distance from the vertex to a point on the edge of the base

Lateral Area as with a pyramid, it is the circumference of the base times the slant height.

Surface Area similar to a pyramid as well, it is the sum of the lateral area and the area of the base.

Theorem 11.4 – Lateral and Surface Areas of a Cone

◦ The lateral area of a right cone is half the product of the circumference of the base and the slant height.

L.A. = · 2Πr ·l -or- L.A. = Πrl

◦ The surface area of a right cone is the sum of the lateral area and the area of the base.

S.A. = L.A. + B

Homework #27Due Thurs/Fri (Apr 11/12)Page 620 – 621◦# 1 – 21 all

Objectives:

To find the volume of a prismTo find the volume of a cylinder

Section 11.4 – Volumes of Prisms and Cylinders

Volume the space that a figure occupies. It is measured in cubic units (, ).

**Notice that volume and surface area are different. Surface area only includes the area

of the container (empty soda can). Volume includes everything inside the container (full

can of soda).**

Theorem 11.5 – Cavalieri’s Principle

◦ If two space figures have the same height and the same cross-sectional area at every level, then they have the same volume.

Theorem 11.6 – Volume of a Prism

◦ The volume of a prism is the product of the area of a base and the height of the prism.

V = B · h

h

B

Theorem 11.7 – Volume of a Cylinder

◦ The volume of a cylinder is the product of the area of the base and the height of the cylinder.

V = B ·h -or- V = Πh

r

hB

Composite Space Figure a 3D figure that is the combination of two or more simpler figures.

◦ Think of a rocket. It is composed of a conical top and a cylindrical body. The composite volume would be the volume of the cone added to the volume of the cylinder.

Objectives:

To find the volume of a pyramidTo find the volume of a cone

Section 11.5 – Volumes of Pyramids and Cones

Theorem 11.8 – Volume of a Pyramid

◦ The volume of a pyramid is one third the product of the area of the base and the height of the pyramid.

V = B · h

Theorem 11.9 – Volume of a Cone

◦ The volume of a cone is one third the product of the area of the base and the height of the cone.

V = B · h -or- V = h

Homework #28Due Monday (April 15)Page 627 – 628◦# 1 – 19 odd

Homework #29Due Monday (April 15)Page 634 – 635◦# 1 – 19 odd

Quiz Tuesday

Objectives:

To find the surface area and volume of a sphere

Section 11.6 – Surface Areas and Volumes of Spheres

Sphere the set of all points in space equidistant from a given point called the center.

Radius a segment that has one endpoint at the center and the other endpoint on the sphere

Diameter a segment passing through the center with endpoints on the sphere

When a plane and a sphere intersect in more than one point, the intersection is a circle. If the center of the circle is also the center of the sphere, the circle is called a great circle.

Circumference of the great circle is the same as the sphere

Hemispheres two equal halves of a sphere. These are created by a great circle.

Theorem 11.10 – Surface Area of a Sphere

◦ The surface area of a sphere is four times the product of Π and the square of the radius of the sphere.

S.A. = 4Π

Theorem 11.11 – Volume of a Sphere

◦ The volume of a sphere is four thirds the product of Π (pi) and the cube of the radius of the sphere.

V = Π

Ex: Earth’s equator is about 24,902 mi long. Approximate the surface area of Earth by finding the surface area of a sphere with circumference 24,902 mi.

Ex: The volume of a sphere is 4200 . Find the surface area to the nearest tenth.

Homework #30Due Monday (April 22)

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Objectives:

To find relationships between the ratios of the areas and volumes of similar solids

Section 11.7 – Areas and Volumes of Similar Solids

Similar Solids have the same shape, and all their corresponding dimensions are proportional.

Similarity Ratio the ratio of corresponding linear dimensions of two similar solids.

**Any two cubes are similar and any two spheres are similar**

Ex: Identifying Similar Solids◦ Are the following figures similar?

3

32

6

6

4

6

5

12 10

Theorem 11.12 – Areas and Volumes of Similar Solids

◦ If the similarity ratio of two similar solids is a : b, then:

◦ 1. The ratio of their corresponding areas is :

◦ 2. The ratio of their volumes is :

Ex: Finding the Similarity Ratio

Find the similarity ratio of two cubes with volumes of 729 and 1331 .

Find the similarity ratio of two similar prisms with surface areas 144 and 324 .

A marble paperweight shaped like a pyramid weighs 0.15 lb. How much

does a similarly shaped marble paperweight weigh if each dimension

is three times as large?

Homework #31Due Monday (April 22)

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Test Thursday/Friday