Unit 11 Surface Area and Volume

• This unit addresses three dimensional objects. • It includes surface area, lateral area, height, slant

height, Euler’s Formula, Platonic Solids, volume, volume of a cone, sphere, pyramid, 3 dimensional objects such as a cube, and a prism.

Standards• SPI’s taught in Unit 11:• SPI 3108.1.1 Give precise mathematical descriptions or definitions of geometric shapes in the plane and space. • SPI 3108.1.3 Use geometric understanding and spatial visualization of geometric solids to solve problems and/or create drawings. • SPI 3108.4.5 Describe solids and/or surfaces in three-dimensional space when given two-dimensional representations for the surfaces of three-

dimensional objects. • SPI 3108.4.9 Use right triangle trigonometry and cross-sections to solve problems involving surface areas and/or volumes of solids. • CLE (Course Level Expectations) found in Unit 11:• CLE 3108.1.1 Use mathematical language, symbols, definitions, proofs and counterexamples correctly and precisely in mathematical reasoning. • CLE 3108.1.7 Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce

accurate and reliable models. • CLE 3108.4.5 Extend the study of planar figures to three-dimensions, including the classical solid figures, and develop analysis through cross-sections. • CLE 3108.4.6 Generate formulas for perimeter, area, and volume, including their use, dimensional analysis, and applications. • CFU (Checks for Understanding) applied to Unit 11:• 3108.1.1 Check solutions after making reasonable estimates in appropriate units of quantities encountered in contextual situations. • 3108.1.5 Use technology, hands-on activities, and manipulatives to develop the language and the concepts of geometry, including specialized

vocabulary (e.g. graphing calculators, interactive geometry software such as Geometer’s Sketchpad and Cabri, algebra tiles, pattern blocks, tessellation tiles, MIRAs, mirrors, spinners, geoboards, conic section models, volume demonstration kits, Polyhedrons, measurement tools, compasses, PentaBlocks, pentominoes, cubes, tangrams).

• 3108.1.7 Recognize the capabilities and the limitations of calculators and computers in solving problems. • 3108.1.10 Use visualization, spatial reasoning, and geometric modeling to solve problems. • 3108.1.11 Identify and sketch solids formed by revolving two-dimensional figures around lines. • 3108.4.1 Recognize that there are geometries, other than Euclidean geometry, in which the parallel postulate is not true and discuss unique properties

of each. • 3108.4.10 Identify and apply properties and relationships of special figures (e.g., isosceles and equilateral triangles, family of quadrilaterals, polygons,

and solids). • 3108.4.23 Describe the polyhedron or solid that can be made from a given net including the Platonic Solids. • 3108.4.24 Develop and use special formulas relating to polyhedra (e.g., Euler’s Formula). • 3108.4.25 Use properties of prisms, pyramids, cylinders, cones, spheres, and hemispheres to solve problems. • 3108.4.26 Describe and draw cross-sections (including the conic sections) of prisms, cylinders, pyramids, spheres, and cones.

Terms

• Polyhedron: a three dimensional figure whose surfaces are polygons

• Face: one side, or surface of a polyhedron (each of these faces is a polygon)

• Edge: a line segment formed by the intersection of two faces

• Vertex: A point where three or more edges intersect• Net: A two-dimensional pattern that you can fold up to

form a three dimensional figure. – One of the simplest nets is the one that forms a cube

Name the 3 dimensional object

• Cube

• Prism

• Regular Pyramid

• Long flat box

Example• Draw a net for this box

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10Top

Back Side

Bottom

Front

Right SideLeft Side

20

20

20

20

10

7

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Euler• Leonhard Paul Euler (15 April 1707 – 18 September 1783)

was a pioneering Swiss mathematician and physicist who spent most of his life in Russia and Germany.

• His surname is pronounced OY-lər in English• Euler made important discoveries in fields as diverse as

calculus and graph theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion of a mathematical function.

• Euler is considered to be the preeminent mathematician of the 18th century and one of the greatest of all time. A statement attributed to Pierre-Simon Laplace expresses Euler's influence on mathematics: "Read Euler, read Euler, he is the master [i.e., teacher] of us all."

Euler’s Formula• The number of faces (F), vertices (V) and

edges (E) of a polyhedron are related by this formula: F+V = E + 2

• Looking at the box to the right calculate the number of edges

• Faces = 6• Vertices = 8• F + V = E + 2• 6 + 8 = E + 2• 14 = E + 2• E = 12

Platonic Solids• A Platonic solid is a convex

polyhedron that is regular, in the sense of a regular polygon.

• Specifically, the faces of a Platonic solid are congruent regular polygons, with the same number of faces meeting at each vertex.

• They have the unique property that the faces, edges and angles of each solid are all congruent

• There are only 5: – Tetrahedron (4 sides)– Hexahedron (6 sides)– Octahedron (8 sides)– Dodecahedron (12 sides)– Icosahedron (20 sides)

http://upload.wikimedia.org/wikipedia/commons/7/70/Tetrahedron.gif

http://upload.wikimedia.org/wikipedia/commons/4/48/Hexahedron.gif http://upload.wikimedia.org/wikipedia/commons/1/14/Octahedron.gif http://upload.wikimedia.org/wikipedia/commons/7/73/Dodecahedron.gif http://upload.wikimedia.org/wikipedia/commons/e/e2/Icosahedron.gif

Surface Areas of Prisms and Cylinders

• Prism: a polyhedron with exactly two congruent, parallel faces called bases

• The other faces on a prism are called lateral faces

• You name a prism by the shape of the bases• Altitude of a prism: the perpendicular

segment that joins the planes of the bases• Height of a prism: The length of an altitude

Example

• This is a pentagonal prism• Base:• Lateral Face• Lateral Edge• Altitude• Height

h

Example

• Triangular Prism• Base:• Lateral Face• Lateral Edge• Altitude• Height

h

Prisms

• A prism may either be a right prism, • or an oblique prism• The book says we canAssumeAll right prisms

Lateral Area/Surface Area

• If you think about it, in a right prism, all lateral faces are rectangles, and all lateral edges are altitudes

• The lateral area of a prism is the sum of the area of all the rectangles (lateral faces)

• The surface area is the sum of the lateral area, and the area of the top and bottom face

Lateral Area• A look at lateral area (the sum of the area of the lateral

faces):• If you look at this prism, there are three rectangles:• To calculate the lateral area, we calculate the area of

each face –in this case there are three, there could be more

• Remember, area is base x height• So we have 8 x 16 + 10 x 16 + 8 x 16 = 416

1616

16

8

8

8

8

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10

1616

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Lateral Area• Instead of adding three separate areas, we can

combine these into one area• Now we have one big rectangle• The base is now 8+10+8 = 26, and the height is 16• So the area is now 26 x 16 = 416 –which is the same• What do we call the base now that it’s combined? • This is called the Perimeter • If you think about it, this is the perimeter of the

triangle

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1616

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Now we have a new equation:Lateral area is Perimeter x Height

Example

• What is the Lateral Area of this triangular prism?

• Lateral Area = Perimeter x height• = (15+12+10) x 20• = 37 x 20 = 740• Now use 25 for the perimeter

and 17 for the height using the calculator…

20

12

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Surface Area• To calculate the surface area, calculate the lateral area,

and add the area of the top and bottom base• Remember

– Area of triangle = ½ x b x h– Area of square = S2

– Area of rectangle = b x h– Area of regular polygon = ½ x a x p –where a = apothem,

and p = perimeter (Apothem is the distance from one side to the center of the polygon)

– THE APOTHEM ON A HEXAGON IS ALWAYS ½ THE SIDE X √3– SO IF YOU HAVE A HEXAGON WITH A SIDE OF 6,

THEN THE APOTHEM IS 3 √3

S = 2aTAN(180/n)A = S/2(Tan(180/n)

Area of a Cylinder• Like a prism, this has 2 bases (the top and the

bottom)• We know how to find the area of a circle (π x r2)• We just need the Lateral Area• If we take the cylinder and unroll it, it looks like

this:• How long is the base of this rectangle?• It is the same as the circumference of the circle,

or 2xrx π –or dx πHH

HH

dx π

Example

• Find the Surface area of this cylinder

• S.A. = L.A. + 2B• L.A. = P x H• P = 2 x r x π = 2 x 4 x π = 25.13• P x H = 25.13 x 23 = 578.05• B = π x r2 = π x 16 = 50.27• S.A. = 578.05 + 2(50.27) = 678.59

23

R = 4

Surface Area• Find the surface area of this

rectangular prism• Surface area = Lateral area + 2b

(where b = area of the base)• Lateral area = p x h• P = 7+7+10+10 = 34• H = 20• Lateral Area = 34 x 20 = 680• B (area of one base) = 10 x 7 = 70• Surface area = L.A. + 2B• Surface Area = 680 + 2(70) = 820

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Assignment

• P. 692 6-17 704 7-19• Worksheet 10-1, 10-3

Surface Area of A Pyramid

• 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 –called the VERTEX of the pyramid

• Pyramids are named after the shape of the base• The altitude of a pyramid is the perpendicular segment from

the base to the vertex –the height is the measurement of this altitude

• A regular pyramid has a base that is a regular polygon AND the faces are congruent Isosceles triangles

• The Slant Height is the length of an altitude of a lateral face

Label the Parts

• Base• Altitude

– Height• Slant Height• Vertex

h

h

Lateral Area• Lateral Area of a pyramid is the sum of the

areas of the congruent lateral faces• Each lateral face is a triangle, where the

base of the triangle is the length of the side of the base

• The height of the triangle is the slant height l

• Therefore the area of one face is ½ x l x s• In this pyramid, there are 4 sides, so it

would be 4(1/2 x l x s)• NOTE 4 x s is the perimeter, so now we can

rewrite this (for ALL pyramids) as Lateral Area = ½ x P x l

• Where P = perimeter, and l = slant height

s

l

Lateral Area and Surface Area

• Lateral area of a pyramid = L.A. = ½ x P x l• Surface area of a pyramid = L.A. + B (where B =

area of the base)

Example• Find the surface area of a square pyramid with

base edges of 5 meters and a slant height of 3 meters

• L.A. = ½ x p x l• L.A. = ½ x (4 x 5) x 3• = ½ x 20 x 3 = 30 m2

• S.A. = L.A. + B• B (area of base) = 5 x 5 = 25 m2

• Surface Area = 30 + 25 = 55 m2

Cones

• A cone has a vertex like a pyramid, but the base is a circle, not a polygon

• The height of cone is the perpendicular distance from the base to the vertex (same as a pyramid)

• The slant height is the distance from the vertex to any point on the edge of the base

Cones

• In a pyramid, the lateral area was ½ the perimeter times slant height l

• A cone is much the same: the lateral area is ½ x 2πr x l This reduces to π x r x l

• The surface area is the Lateral Area L.A. + B (area of base) –and the area of the base is

• πr2

Example

• Find the surface area of a cone with a slant height of 25 cm, and a radius of 15 cm in terms of π

• L.A. = π x r x l• L.A. = π x 15 x 25= 375π• S.A. = L.A. + B (area of base)• S.A. = 375 π + 152π • S.A. = 375π + 225π = 600π

Assignment

• Text Page P. 713 9-21• Worksheet 10-4

Spheres

• A sphere (really a ball) is the set of all points which are equally distant from a center.

• A radius is the same as in a circle, it starts at a point on the edge, and ends at the center point

• A diameter is also the same –it is a segment that ends on both sides of the sphere, and goes through the middle

Sphere• Imagine cutting through a sphere with a

plane. If you took each half and looked at it, it would create a circle

• If you cut through the sphere exactly at the center, the circle you create would be called the GREAT CIRCLE

• This Great Circle is the circumference of the sphere –it is the largest –or longest- distance around the sphere

• A Great Circle creates two hemispheres (hemi means half)

S.A. –The surface area of a sphere is calculated as 4πr2

Volume: 4/3πr3

Example

• Find the surface area of a sphere with a diameter of 8m. Leave in terms of π

• Surface Area = 4πr2

• = 4 π42

• = 64 π m2

Assignment

• Page 737 6-25• Worksheet 10-7

Volume of Prisms and Cylinders

• We know that volume in its most simple terms is length x width x height

• We can refine this somewhat and say that for prisms, volume is the area of the base x the height

• V = Bh• Here, the area of the base is

6 x 4 = 24. Then multiply times the height (12) = 288 6

4

12

Example

• Find the volume of this triangular prism

• Here, we must find the area of the base triangle

• A = ½(bxh) = ½(8x8) = ½(64) = 32

• Multiply x height• = 32 x 24 = 768

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Volume of a Cylinder

• We use the same equation for the volume of a cylinder

• V = Bh –(where B = the area of the base) x height

• In this example, the area of the base is• A = πr2= πx9 = 9π• V = 9π x 8 = 72π or 226.2

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8

Example• Find the volume of this composite figure• Break this down into two figures, a rectangle and

a cylinder • What is the diameter of the circle on top?

– 12 cm• What is the radius of the circle on top?

– 6 cm• Therefore, what is the height of the rectangle?• 17 cm – 6 cm = 11 cm• The volume of the rectangle is B x h = 16 x 12 x 17

= 3264• The volume of the cylinder is B x h• B = πr2 = π x 62 = 36π• Volume = 36π x 16 = 1809.6

16 cm 12 cm

17 c

m

But we only need ½ the cylinder, so we have 1/2x(1809.6) = 904.8Now add 3264 = 4168.8 cm3

Example

• Find the volume of this composite figure• There is a cube, which is 8 x 8 x 8 = 512• And a rectangular prism which is 8 x 16 x 8 = 1024• Total = 512 + 1024 = 1536

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Assignment

• Text Page 721 6-16• Worksheet 10-5

Volume of Pyramids and Cones

• The volume of a pyramid is 1/3 x B x h (note, this is NOT slant height, but regular height)

• Find the volume of this pyramid• V = 1/3 x (6 x 8) x 25• V = 400 25

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8

Volume of a Cone

• The volume of a cone is the same as the volume of a sphere

• V = 1/3 x B x h• In this example, • V = 1/3 x (π) x 62 x 14• V = 527.8

H = 14

R = 6

Assignment

• Page 730 9-19• Worksheet 10-6