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# Ch 4 - PowerPoint PPT Presentation

Ch 4 . Measuring Prisms and Cylinders . Area of a Rectangle . To find the area of a rectangle, multiply its length by its width. A = l x w. How do you find the area. So, the area is 8.36 cm 2 . Area of a Triangle .

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### Ch 4

Measuring Prisms and Cylinders

• To find the area of a rectangle, multiply its length by its width.

• A = l x w

• So, the area is 8.36 cm2.

• To find the area of a triangle, multiply its base by its height, then divide by 2.

• Remember the height of a triangle is perpendicular to its base.

• The formula for the area of a triangle can be written:

How do you find the area

Substitute b = 12 and h = 3.

So, the area is 18 m2.

Watch Brainpop: Area of Polygons

• http://www.brainpop.com/math/geometryandmeasurement/areaofpolygons/preview.weml

• To find the area of a circle, use the formula:

• A = r 2

• where r represents the radius of the circle

• Recall that  is a non-terminating and a non-repeating decimal number. So, any calculations involving  are approximate.

• You need to use the  function on your calculator to be more accurate – 3.14 is not accurate enough.

• Use A = r 2.

Substitute r = 6 ÷ 2 = 3.

So, the area is about 28 mm2.

• The perimeter of a circle is named the circumference.The circumference is given by: C = d or C = 2r (Recall: d = 2r)

The circumference of the circle is about 22 cm.

• The circumference of a circle is 12.57 cm

• To find the radius of the circle, divide the circumference by 2.

Watch Brainpop: Circles

• http://www.brainpop.com/math/geometryandmeasurement/circles/preview.weml

• Workbook pg 74 – 75

• Puzzle Package

Watch Brainpop: Review

• http://www.brainpop.com/math/numbersandoperations/pi/preview.weml

• http://www.brainpop.com/math/geometryandmeasurement/polygons/preview.weml

• http://www.brainpop.com/math/geometryandmeasurement/polyhedrons/preview.weml

• http://www.brainpop.com/math/geometryandmeasurement/typesoftriangles/preview.weml

### 4.1 Exploring Nets

• A prism has 2 congruent bases and is named for its bases.

• When all its faces, other than the bases, are rectangles and they are perpendicular to the bases, the prism is a right prism.

• A regular prism has regular polygons as bases.

• A regular pyramid has a regular polygon as its base. Its other faces are triangles. They are named after its base.

Regular OctagonalPyramid

Regular Pentagonal Pyramid

Regular Square Pyramid

• ACube

• 6 congruent squares

• A Right Square Pyramid

• 1 square, 4 congruent isosceles triangles

• A Right Pentagonal Prism

• 2 regular pentagons, 5 congruent rectangles

• A net is a diagram that can be folded to make an object. A net shows all the faces of an object

• http://www.senteacher.org/wk/3dshape.php

• A net is a two-dimensional shape that, when folded, encloses a three-dimensional object. http://www.youtube.com/watch?v=9KXuaT18Jyw&feature=related (quick)

• The same 3-D object can be created by folding different nets.

• You can draw a net for an object by visualizing what it would look like if you cut along the edges and flattened it out.

### 4.2 Creating Objects From Nets

This diagram is not a rectangular prism!

Here is the net being put together

p. 178 of TextbookExample #1a

2 congruent regular pentagons

5 congruent rectangles

p. 178 of TextbookExample #1b

1 square

4 congruent isosceles triangles

p. 178 of TextbookExample #1c

This net has 2 congruent equilateral triangles and 3 congruent rectangles.

The diagram is a net of a right triangular prism. It has equilateral triangular bases.

p. 178 of TextbookExample #1d

This is not a net. The two triangular faces will overlap when folded, and the opposite face is missing.

Move one triangular face from the top right to the top left. It will now make a net of an octagonal pyramid.

p. 179 of TextbookExample #2

### 4.3 Surface Area of a Right Rectangular Prism

• Surface area is the number of square units needed to cover a 3D object

• It is the sum of the areas of allthe faces of an object

• The SA of a rectangular prism is the sum of the areas of its rectangular faces. To determine the surface area of a rectangular prism, identify each side with a letter.

• Rectangle A has an area of

• A = l x w

• A = 4 x 5

• A = 20

• Rectangle B has an area of

• A = l x w

• A = 7 x 5

• A = 35

• Rectangle C has an area of

• A = l x w

• A = 7 x 4

• A = 28

C

B

A

• To calculate surface area we will need 2 of each side and add them together

• SA = 2(A) + 2(B) + 2(C)

• SA = 2(l x w) + 2(l x w) + 2(l x w)

• SA= 2(4 x 5) + 2(7 x 5) + 2(4 x 7)

• SA = 2(20) + 2(35) + 2(28)

• SA = 40 + 70 + 56

• SA = 166 in2

C

B

A

• Another Example

• SA = 2(A) + 2(B) + 2(C)

• SA = 2(l x w) + 2(l x w) + 2(l x w)

• SA= 2(8 x 10) + 2(7 x 8) + 2(10 x 7)

• SA = 2(80) + 2(56) + 2(70)

• SA = 160 + 112 + 140

• SA = 412 units2

C

B

A

• You Try

• SA = 2(A) + 2(B) + 2(C)

• SA = 2 (l x w) + 2(l x w) + 2 (l x w)

• SA= 2(15 x 6) + 2(10 x 6) + 2(10 x 15)

• SA = 2(90) + 2(60) + 2(150)

• SA = 180 + 120 + 300

• SA = 600 cm2

C

B

A

• Pg p.186

• #4, 6,7,10,12,13,15

### 4.4 Surface Area of a Right Triangular Prism

• To calculate the surface area of right triangular prism, draw out the net and calculate the surface area of each face and add them together.

• Draw and label the net

20cm

16cm

D

16cm

20cm

12cm

10cm

B

10cm

C

10cm

A

D

16cm

20cm

• SA = A area + B area + C area + 2 D area

• SA = A(l x w) +B(l x w) + C(l x w) + 2 * D[(b x h) ∕ 2)]

• SA = A(16 x 10) + B(10 x 12) + C(20 x 10) + 2 * D[(16 x 12) ∕ 2]

• SA = 160 + 120 + 200 + (2 * 96)

• SA = 160 + 120 + 200 + 192

• SA = 672 cm2

20cm

16cm

D

16cm

20cm

12cm

10cm

B

10cm

C

10cm

A

D

16cm

20cm

• Draw and label the net

10cm

6cm

D

6cm

10cm

8cm

15cm

B

15cm

C

15cm

A

D

6cm

10cm

• SA = A area + B area + C area + 2 D area

• SA = A(l x w) + B(l x w) + C(l x w) + 2 * D[(b x h) ∕ 2)]

• SA = A(15 x 6) + B(15 x 8) + C(15 x 10) + 2 * D[(6 x 8) ∕ 2]

• SA = 90 + 120 + 150 + (2 * 18)

• SA = 90 + 120 + 150 + 36

• SA = 396 cm2

10cm

6cm

D

6cm

10cm

8cm

15cm

B

15cm

C

15cm

A

D

6cm

10cm

• Draw and label the net

2.7m

1.4m

D

1.4m

2.7m

2.3m

0.7m

B

0.7m

C

0.7m

A

D

1.4m

2.7m

• SA = A area + B area + C area + 2 D area

• SA = (la x wa) + (lb x wb) + (lc x wc) + 2[(ld x wd) ∕ 2)]

• SA = (1.4 x 0.7) + (0.7 x 2.3) + (2.7 x 0.7) + 2 [(1.4 x 2.7) ∕ 2]

• SA = 0.98 + 1.61 + 1.89 + 2(1.89)

• SA = 8.26m2

2.7m

1.4m

D

1.4m

2.7m

2.3m

0.7m

B

0.7m

C

0.7m

A

D

1.4m

2.7m

• Draw and label the net

? m

3m

D

3m

?m

8cm

7 m

B

7 m

7 m

C

7 m

A

D

3cm

?m

• Before you can solve, you need to find the missing side using Pythagorean Theorem!

? m

3m

D

3m

?m

8cm

7 m

B

7 m

C

7 m

A

D

3cm

?m

• Pythagorean Theorem

• a2 + b2 = h2

• 32 + 82 = h2

• 9 + 64 = h2

• 73 = h2

• √73 = √ h2

• 8.54 = h

8.54m

3m

8.54m

3m

8m

7 m

7 m

7 m

3m

?m

• SA = A area + B area + C area + 2 D area

• SA = A(l x w) + B(l x w) + C(l x w) + 2 * D[(l x w) ∕ 2)]

• SA = A(3 x 7) + B(7 x 8) + C(8.54 x 7) + 2 * D[(3x 8) ∕ 2]

• SA = 21 + 56 + 59.78 + (2 * 12)

• SA = 21 + 56 + 59.78 + 24

• SA = 160.78 m2

8.54m

3m

D

8.54m

3m

8m

7 m

B

C

7 m

7 m

A

D

3m

?m

• P. 191

### 4.5 Volume of a Right Rectangular Prism

• VIDEO Volume of a right rectangle

• V = Ah

• Volume equals the area of the base of the rectangle times its height – this applies to all right cylinders or prisms

• V = Ah

• V = (l x w) x h

• V = (13.5 x 5) x 18.5

• V = 67.5 x 18.5 (not really needed)

• V = 1248.75 cm3

Jack’s family opened a full carton of frozen yogurt for dessert. After they ate, there was ¾ left. Jack wants to know what volume of frozen yogurt they ate. He measured the carton to be to have a length of 12cm, a width or 9 cm and a height of 18cm.

• V = Ah

• V = (l x w) x h

• V = (12 x 9) x 18

• V = 1944 cm3 (Volume of entire container – we want ¾ of the container )

• V = 1944 x ¾

• V = 1458 cm3

• Read “Quick Review” on pg. 85

• Complete pg 85-86

### 4.6 Volume of a Right Triangular Prism

• Relationship of a right triangular prism with right rectangular prism

• Volume is equal to the area of the base of the triangle times the length

• V = Al

• V = Al

• V = (b * h /2) x l

• V = (3 x 8 /2) x 7

• V = (24 /2) x 7

• V = 12 x 7

• V = 84 m3

• A glass vase in the shape of a right triangular prism is filled with colouredsand as a decoration. What is the volume of the vase. It measured 15cm by 8cm by 80cm

• V = Al

• V = (b * h /2) x l

• V = (15 x 8 /2) x 80

• V = 60 x 80

• V = 4800 cm3

Watch Brainpop: Volume of Prisms

• http://www.brainpop.com/math/geometryandmeasurement/volumeofprisms/preview.weml

• Complete pg. 87 - 89

### 4.7 Surface Area of a Right Cylinder

• Imagine a can. If you unwrap the label off the can, the lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

Notes lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• To find the surface area of a cylinder – sketch the net

• Remember that the width of the rectangle will be equal to the circumference of the circle (2πr)

• SA = 2(area of a circle) + area of the rectangle

Example 1 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

SA = 2πr2 + (2πr)(h)

SA = 2π(3.1)2 + (2π(3.1))(12)

SA = 2π(9.61)+ (2π(3.1))(12)

SA = 120.762822 + 230.734493

SA = 354.50inches2

Example 2 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

SA = 2πr2 + (2πr)(h)

SA = 2π(9)2 + (2π(9))(11)

SA = 2π(81)+ (2π(9))(11)

SA = 508.938010 + 622.035345

SA = 1130.97m2

11m

9m

Example 3 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• Calculate the outside surface area of the cylinder. The cylinder is open on one end.

SA = πr2 + (2πr)(h)

SA = π(3)2 + (2π(3))(11)

SA = π(9)+ (2π(3))(11)

SA = 28.274333 + 207.345115

SA = 235.62m2

8m

3m

Review Video lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

Workbook lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• Complete pg 90 - 92

### 4.8 Volume of a Right Cylinder lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

The Formula lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• Cylinders follow the same properties as prisms for volume

• V = area of the base x height

• V = πr2 x h

Example 1 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

V = πr2 x h

V = π(22)x 6

V = π(4) x 6

V = 12.6 x 6

V = 75.6 m3

6m

2m

Example 2 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• Martha has a choice of two different popcorn containers at a movie. Both containers are the same price. Which container should Martha buy is she wants more popcorn for her money.

• Container 1: has a diameter of 20cm and height of 40cm.

• Container 2 has a diameter of 30cm and a height of 20cm.

Container 1 lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

Container 2

Radius is half the diameter so r = 15cm

V = πr2 x h

V = π(152)x 20

V = π(225) x 20

V = 706.9 x 20

V = 14138 cm3

• Radius is half the diameter so r = 10cm

V = πr2 x h

V = π(102)x 40

V = π(100) x 40

V = 314.2 x 40

V = 12568 cm3

Watch lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.Brainpop: Volume of Cylinders

• http://www.brainpop.com/math/geometryandmeasurement/volumeofcylinders/preview.weml

Workbook lateral surface, or label, would be a rectangle. The height of the label is the same height as the can. The length of the label is the diameter of the base of the can because the label wraps all the way around the can.

• Complete pg 93 - 94