Math 6: Geometry Notes 2-Dimensional and 3-Dimensional …rpdp.net/admin/images/uploads/resource_6228.pdf · 2019-04-21 · Math 6 Notes Geometry: 2-Dimensional and 3-Dimensional

Math 6 Notes Geometry: 2-Dimensional and 3-Dimensional Figures Page 1 of 46 Revised 2014-NACS

Math 6: Geometry Notes

2-Dimensional and 3-Dimensional Figures

Short Review: Classifying Polygons

A polygon is defined as a closed geometric figure formed by connecting line segments endpoint

to endpoint.

Polygons are named by the number of sides. We know a triangle has 3 sides. Below are the

names of other polygons.

Short Review: Classifying Triangles

Triangles can be classified by the measures of their angles:

acute triangle—3 acute angles

right triangle—1 right angle

obtuse triangle—1 obtuse angle

Example: Classify each triangle by their angle measure:

Acute (Equiangular) Right Obtuse Acute

Triangles can also be classified by the lengths of their sides. You can show tick marks to show

congruent sides.

equilateral triangle—3 congruent sides

isosceles triangle—at least 2 congruent sides

scalene triangle—no congruent sides

Polygons Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon

# of sides 4 5 6 7 8 9 10

Polygons Not Polygons

40°

L

A M

80° 60° 50°

20°

60° 45° E

D

F

G

H J C B K

40° 60°

120°

55°


2.5 10 15

12.5

1

5

2.5

x

x

x

Example: Classify the triangle. The perimeter of the triangle is 15 cm.

Using the information given regarding the perimeter:

Since 2 sides are congruent, the triangle is isosceles.

A tree diagram could also be used to show the triangle relationships.

Short Review: Classifying Quadrilaterals

A quadrilateral is a plane figure with four sides and four angles. They are classified based on

congruent sides, parallel sides and right angles.

Quadrilateral Type Definition Example

Parallelogram Quadrilateral with both pairs of

opposite sides parallel.

Rhombus Parallelogram with four

congruent sides.

Rectangle Parallelogram with four right

angles.

10 cm

2.5 cm

x

Tree Diagram for Triangles

triangles

acute obtuse right

scalene isosceles

equilateral

scalene isosceles

scalene isosceles

Note: This

polygon is a

parallelogram.

Note: This polygon is a

parallelogram.

>>

>>

equilateral isosceles scalene


Square Parallelogram with four right

angles and four congruent sides.

Trapezoid Quadrilateral with exactly one

pair of parallel sides.

Another way to show the relationship of the parallelograms is to complete a Venn diagram as

shown below.

Vocabulary becomes very important when trying to solve word problems about quadrilaterals.

Example: A quadrilateral has both pairs of opposite sides parallel. One set of opposite angles

are congruent and acute. The other set of angles is congruent and obtuse. All four

sides are NOT congruent. Which name below best classifies this figure?

A. parallelogram

B. rectangle

C. rhombus

D. trapezoid

We have both pairs of opposite sides parallel, so it cannot be the trapezoid. Since the angles

are not 90° in measure, we can rule out the rectangle. We are told that the 4 sides are not

congruent, so it cannot be the rhombus. Therefore, we have a parallelogram. (A)

6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons

by composing into rectangles or decomposing into triangles and other shapes; apply these

techniques in the context of solving real-world and mathematical problems.

Note: This

polygon is a

parallelogram.

>>

>>

Squares

Parallelograms

Rectangles Rhombi


Area of Triangles and Quadrilaterals

One way to describe the size of a room is by naming its dimensions. A room that measures 12 ft.

by 10 ft. would be described by saying it’s a 12 by 10 foot room. That’s easy enough. There is nothing wrong with that description. In geometry, rather than talking about a room, we

might talk about the size of a rectangular region.

For instance, let’s say I have a closet with dimensions 2 feet by 6 feet (sometimes given as 2 6 ).

That’s the size of the closet. Someone else might choose to describe the closet by determining how many one foot by one foot

tiles it would take to cover the floor. To demonstrate, let me divide that closet into one foot

squares.

By simply counting the number of squares that fit inside that region, we find there are 12

squares. If I continue making rectangles of different dimensions, I would be able to describe their size by

those dimensions, or I could mark off units and determine how many equally sized squares can

be made.

Rather than describing the rectangle by its dimensions or counting the number of squares to

determine its size, we could multiply its dimensions together.

Putting this into perspective, we see the number of squares that fits inside a rectangular region is

referred to as the area. A shortcut to determine that number of squares is to multiply the base by

the height.

The area of a rectangle is equal to the product of the length of the base and the length of a

height to that base.

That is . Most books refer to the longer side of a rectangle as the length (l), the shorter

side as the width (w). That results in the formula . The answer in an area problem is

always given in square units because we are determining how many squares fit inside the region.

2 ft.

6 ft.

2 ft.

6 ft.


3 2A

1

3 9

9 3

3

yards x

feet

x

x

3 2A

Example: Find the area of a rectangle with the dimensions 3 m by 2 m.

The area of the rectangle is 6 m2.

Example: Find the area of the rectangle.

Be careful! Area of a rectangle is easy to find, and students may quickly multiply to

get an answer of 18. This is wrong because the measurements are in different units.

We must first convert feet into yards, or yards into feet.

We now have a rectangle with dimensions 3 yd. by 2 yd.

The area of our rectangle is 6 square yards.

If I were to cut one corner of a rectangle and place it on the other side, I would have the

following:

A parallelogram! Notice, to form a parallelogram, we cut a piece of a rectangle from one side

and placed it on the other side. Do you think we changed the area? The answer is no. All we

did was rearrange it; the area of the new figure, the parallelogram, is the same as the original

rectangle.

This allows us to find a formula for the area of a parallelogram.

2 yd.

9 ft.

base

hei

ght

hei

ght

base


3 6

18

A

A

Since the bottom length of the rectangle was not changed by

cutting, it will be used as the base length (b), the height of the

rectangle was not changed either, we’ll call that h.

Now we arrive at the formula for the area of a parallelogram.

.

Example: The height of a parallelogram is twice the base. If the base of the parallelogram is 3

meters, what is its area?

First, find the height. Since the base is 3 meters, the height would be twice that or 2(3) or 6 m.

To find the area,

The area of the parallelogram is 18 m2 .

We have established that the area of a parallelogram is . Let’s see how that helps us to

understand the area formula for a triangle and trapezoid.

Remember: Once a formula for a figure has been developed, it can be used for any figure that

meets its criteria.

For example: The parallelogram formula can be used for rectangles, rhombi, and squares.

The rectangle formula can be used for squares.

The rhombus formula (derived in HS Geometry) can be used for squares.

This is based on the Venn Diagram given previously (pg. 3 of these notes). The inner sets have

all the same attributes and properties of the sets they are contained within. Therefore, what must

be true about any element of the outer set must be true of all elements of that set.

b

h


Composite Figures are figures made up of multiple shapes. (linkage - Composite numbers have

multiple factors) In order to find the area of these oddly-shaped figures they must be

decomposed into figures we are familiar with.

Let’s start with the trapezoid…

Of course the trapezoid formula can be used but we can also decompose this trapezoid into a

rectangle and two triangles. The area of this trapezoid would be the area of the rectangle added to

the areas of the two triangles.

For this parallelogram, its base is 8 units

and its height is 2 units. Therefore, the

area is .

h

base If we draw a line strategically, we can cut

the parallelogram into 2 congruent

trapezoids. One trapezoid would have an

area of one-half of the parallelogram’s area

(8 units2).

Height remains the same. The

base would be written as the sum of

. For a trapezoid:

h

base

base

h

If we draw a diagonal, it cuts the

parallelogram into 2 triangles. That means

one triangle would have one-half of the

area or 6 units2. Note the base and height

stay the same. So for a triangle,

base

h

For this parallelogram, its base is 4 units

and its height is 3 units. Therefore, the

area is .

10 in.

4 in.

6 in.


In this case, because the trapezoid is isosceles, the two

triangles will be congruent

Notice the 1b length 10 is equal to the

2b length 6, plus 4

more inches. Those 4 inches can be split into 2 and 2 and

would indicate the lengths of the bases of the two triangles.

Now we can use the rectangle formula and triangle formula (twice) to find the total area.

Rectangle Triangle

6 4

24A

A

2

4

1

2

14

2

A

A

Since the two triangles are congruent,

3

24

2

2 4

A

A

The area of this trapezoid is 32 square

inches.

By applying the trapezoid formula we can check to see if our answer is correct.

110 6 4

2

116 4

32

2

A

A

A

Once again, we see the area of the trapezoid is 32 in2.

Now that we see that it can be done, we can explore the areas of other, less common, composite

shapes. These shapes must be decomposed and the area formulas of the decomposed parts can be

used to find each individual area. The total area of the figure is the sum of the areas of its

decomposed parts.

10 in.

4 in.

6 in.

6 in.

4 in.

6 in.

2 in. 2 in.


Example: Find the area of the given shape.

Start by decomposing…

One way to decompose is given above. Students may find multiple ways to decompose the given

figures. If the needed measurements are not available for them to complete the problem, they

may want to consider trying a different combination. (Re-decompose?)

From the diagram we can see that the total area of this figure will be the sum of the areas of the

three rectangles that composed it. Notice that the measures of the sides of the figure can be found

by counting the blocks that run along the side of that portion. (Be careful of the scale of the

diagram, sometimes a block represents more than one unit.)

The area of this figure is 45 square cm.

Students can verify this answer by counting the

squares inside the figure.

Example: Find the area of the given polygon. (DOK 2)

First, the figure must be decomposed…

Students should be able to find a rectangle and a triangle.

A. 102 m2

B. 187 m2

C. 289 m2

D. 391 m2

So, the area of this figure is 289 m2. (C)

A B

C

7cm-5cm = 2cm

7 3 2 3 6 3

21 6 8

5

1

4A

A

A

TOTAL

remember... and

111 17 17 12

2

187

1

289

2

102

lw

A

b

A

A

h

TOTAL

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Example: Find the area of the given polygon. (DOK 2)

A. 45 cm

2

B. 90 cm2

C. 110 cm2

D. 180 cm2

This figure can be decomposed a few ways:

Vertically down the middle, forming two congruent triangles:

b = 11 + 4 = 15 cm, h = 12 2 = 6 cm

Therefore, the total area is twice the area of one triangle.

1

2

12

12 15

2

22

90

6TOTAL

h

A

bh

b

A

The total area is 90 cm

2. (b)

Another way would be:

Horizontally, along the diagonal, forming two non-congruent triangles:

b = 12 cm (on both), h1 = 4 cm (top ∆) & h2 = 11 cm (bottom ∆)

In this case we must find the sum of both areas to find the total sum.

1 1

12 4 12 112 2

24 66

90

1

2

TOTAL

TOTAL

A

A

A

bh

The total area is 90 cm2. (b)


A final alternative would be:

Horizontally and vertically, forming four right triangles, with two sets

congruent: (these measurements are interchangeable based on your

perspective when viewing the triangles)

b1 = 4 cm (both smaller ∆s) & b2 = 11 cm (both larger ∆s),

h = 12 2 = 6 cm

This time we must add the areas of all four triangles together, but recall that there were two sets

of congruent triangles formed when we decomposed the original figure in this manner. So…

We can find the area of one small triangle and double it, then find the area of one larger triangle

and double it, and finally add those two doubled areas together.

1

2

1 12 4 6 2 11 6

2 2

24 6

0

2

9

6

2

TOTAL

TOTAL

A

A

A

bh

Taa daa…The total area is 90 cm2. (b) AGAIN!!

Example: Find the area of the given polygon.

The area of this figure can be found multiple ways.

It could be decomposed into rectangles this way…

The total area is 14.96 cm2.

An alternative to this method and extension on this topic is to decompose and subtract. In this

case, students can picture an imaginary rectangle surrounding the entire figure then subtract the

region that is not a part of the original figure.

1.2 6.4 3.6 0.6 0.8 6.4

7.68 2.16

1

5 1

4.9

.

6

2

A

A

A

TOTAL


Again, the total area is 14.96 cm2.

Extensions Examples: Find the areas of the shaded regions in the figures below.

a)

The area of the shaded region is 12 square cm.

b)

The area of the shaded region is 60 square cm.

Three-Dimensional Figures

A solid is a three-dimensional figure that occupies a part of space. The polygons that form the

sides of a solid are called a faces. Where the faces meet in segments are called edges. Edges

meet at vertices.

A prism is a solid formed by polygons. The faces are rectangles. The

bases are congruent polygons that lie in parallel planes.

5.6 6.4 3.6 5.8

3

14.96

5.84 20.88A

A

A

TOTAL

All squares are rectangles...

1

4 2

16

2

4 2

4

A

A

A

TOTAL

To find the area of the inner region, we must decompose it.

4 2 2 2

8 4

Now the shaded region can be fo

12

und

inner regionA

A

A

.

60

8 9 12

72 12

A

A

A


A pyramid is a solid whose base may be any polygon, with the other

faces triangles.

Polyhedra are solids with all faces as polygons. Prisms and pyramids would meet this criterion,

while cylinders and cones would not, therefore they will be discussed at a later time.

“A picture is worth a thousand words.”

The ability to draw three-dimensional figures is an important visual thinking tool. Here are

some drawing tips:

Rectangular Prism (face closest to you):

Draw the front

rectangle.

Draw a congruent

rectangle in another

position.

Connect the corners

of the rectangles.

Use dashed lines to

show the edges you

would not see.

Your rectangular

prism!

Rectangular Prism (edge closest to you):

6.G.A.2 Find the volume of a right rectangular prism with fractional edge lengths by packing

it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume

is the same as would be found by multiplying the edge lengths of the prism. Apply the

formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional

edge lengths in the context of solving real-world and mathematical problems.

Volume

If you were to buy dirt for your yard, it’s typically sold in cubic yards—that’s describing volume.

If you were laying a foundation for a house or putting in a driveway, you’d want to buy cement,

and cement is often sold by the cubic yard. Carpenters, painters and plumbers all use volume

relationships.

base

vertex


The volume of a three dimensional figure measures how many cubes will fit

inside it. It’s easy to find the volume of a solid if it is a rectangular prism

with whole number dimensions. Let’s consider a figure 3 m 2 m 4 m.

We can count the cubes measuring 1 meter on an edge. The bottom layer is

3 2—there are 6 square meter cubes on the bottom layer.

We have three more layers stacked above it (for a total of 4 layers), or 6 + 6

+ 6 + 6 = 24.

Now we can reason that if I know how many cubes are in the first layer (6),

then to find the total number of cubes in the stack, you simply multiply the

number on the first layer by the height of the stack 6 4 24 .

This is a way of finding volume. We find the area of the base (B) and multiply it times the

height (h) of the object.

For prisms, , .where B is the area of the base and h is the height Since rectangular prisms have bases that are rectangles, .

Therefore, we use the formula .

The answer in a volume problem is always given in cubic units (cm3, in

3, ft

3,…) because we are

determining how many cubes will fill the solid.

For testing at the state level (grade 6) on volume, solid figures may only include cubes or

right rectangular prisms. For that reason, the focus for volume should be .

Example: Julie is using sugar cubes to create a model for a school project; each sugar cube has

an edge length of 1 cm. After building her first model, she realizes that she must increase

each measure by 1½ times. The diagram given shows her first try.

a) How many cubes did she use to build her first model?

3 4

24

2

V

V

The volume is 24 cm3.

b) What would the new dimensions be after she increases each measure?

Original Measure Rate of Change →

(slope) & (dilations)

New Measure

Length = 3 cm 1½ 31½ = 4½ cm

Width = 2 cm 1½ 21½ = 3 cm

Height = 4 cm 1½ 41½ = 6 cm

3 m

4 m

2 m

1 m

1 m

3 m 2 m

3 m

4 m

2 m


c) How many more sugar cubes will she need to complete her project?

First, she must find the volume of the larger model, then subtract the volume

of the original model to find the number of cubes she will need to finish.

This could be done physically… by adding two blocks on top (height), 1

block behind (width) and 1½ blocks to the side (length).

OR

Just use the formula…

The larger volume is 81 cm3. 81-47=57, so

she needs 57 more sugar cubes.

Example: The diagram below shows a cube with sides of length 30cm. A smaller cube with side

length 5 cm has been cut out of the larger cube.

a) What is the volume of the large cube before the small cube is cut out?

The volume of the

large cube is 27,000 cm3.

b) What is the volume of the small cube being cut out?

The volume of the small cube is 125 cm3.

c) What is the volume of the solid left?

The total volume of the

solid is 26,875 cm3.

3 m

4 m

2 m

1

4

81

3 62

V

V

400 780 260 360 342 342

2484

S

V

A

5 5

125

5V

V

27,000 125

26,875


Example: If the volume of the rectangular prism is 450,000 cm3, the value of x is …

a) 0.06 m

b) 0.6 m

c) 6 m

d) 60 m

Note the units in the question and the units used in the answer choices. We need to

convert so we are comparing like measurements. Every meter contains 100 centimeters

so a cubic centimeter would measure 100 100 100cm cm cm or 31,000,000cm , therefore

the volume into m3…(converting using ratios)

33 3

3

1450,000 0.45

1,000,000

mcm m

cm and 0.45 1.5 0.5

0.45 0.75

x

x

.

Students can guess and check to find the answer. Some number sense can make the

answer almost obvious. 0.45 is a little more than half of 0.75, so to find the answer 0.75

must be multiplied by a little more than half. Only one choice fits that criterion… (b)

Example: By how much will the volume of a rectangular prism increase, if its length, width, and

height are doubled?

a) 4 times

b) 2 times

c) 6 times

d) 8 times

Students can use a strategy of looking for a pattern…

Volume of a 1 1 1 cube 1

1

1 1

V

V

= 31unit ,

now double the measures…


8

2 2

V

V

= 38units WOW, 8 times as much. (d)

Example: By how much will the volume of a rectangular prism increase, if its length is

doubled?

a) 2 times

b) 4 times

c) 8 times

d) 6 times


Again, students can look for a pattern…


1

1 1

V

V

= 31unit ,

now double the length measure…


2

1 1

V

V

= 32units 2 times as much this time. (a)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use

coordinates to find the length of a side joining points with the same first coordinate or the

same second coordinate. Apply these techniques in the context of solving real-world and

mathematical problems.

Example: Find the length of segment CD. (DOK 1)

Simply count the units along the indicated side.

Common error…

Remind students to start at “0” and count each “swoop” as 1

unit.

CD = 3 units

Example: In each part below two sides of a rectangle are shown. Write the coordinates of the

fourth corner of each rectangle. Then answer the questions.

a)

The fourth corner is at (5, -5).

Can the perimeter and area of this

rectangle be found?

If so, what are they?

If not, why not?

Yes, they are…

2 10 2 9

38

20 18

P

P

P

and 10

90

9A

A

The perimeter is 38 units.

The area is 90 units2.

1 2 3


12

4 3

P

P

b)

The fourth corner is at (5, 0).

Can the perimeter and area of this

rectangle be found?

If so, what are they?

If not, why not?

NO, because the sides are not horizontal or

vertical so the measures cannot be

determined.

Example: Which gives the perimeter of the square? (DOK 2)

By counting the units we can see that l = 3 &

w = 3, therefore…

2 3 2 3

6 6

12

P

P

P

OR

Since the side lengths of a square are always equal, it may be

faster for students to find one side length and multiply by 4.

Both methods yield the answer 12 units. (d)

Example: Plot the sixteen points in the table below on this graph. After graphing the points,

connect them to make a 16-pointed star.

POINTS POINTS POINTS POINTS

A(4, 0) E(-4, 0) I(0, 4) M(0, -4)

B(1, 2) F(1, -2) J(3, 3) N(3, -3)

C(2, 1) G(2, -1) K(-1, 2) P(-1, -2)

D(-3, 3) H(-3, -3) L(-2, 1) Q(-2, -1)

a) 4 c) 10

b) 8 d) 12


a) Find all horizontal lengths:

DJ = __6__ units KB = __2__ units LC = __4__ units

EA = __8__ units QG = __4__ units PF = __2__ units

HN = __6__ units

b) Find all vertical lengths:

DH = __6__ units LQ = __2__ units KP = __4__ units

IM = __8__ units BF = __4__ units CG = __2__ units

JN = __6__ units

c) Find the area of the triangle found by connecting the vowels (∆AEI)

1

8 4

2

16

2

1bh

A

A

The area of ∆AEI is 16 square units or 16 units2.

d) Discussion: I purposely skipped labeling any point with an “O”, what point is usually

labeled with an “O”? Give its name and coordinates. The origin, (0, 0)

Example: In each question below the coordinates of three corners of a square are given. Find

the coordinates of the other corner in each case. You may find it helpful to draw a sketch.

a) (2, -2), (2, 3) and (-3, 3). The other corner of the square is at ( ___, ___). (-3, -2)

b) (2, 3), (3, 4) and (1, 4). The other corner of this square is at ( ___, ___). (2, 5)

c) (2, 2), (4, 4) and (4, 0). The other corner of this square is at ( ___, ___). (6, 2)

A

B C

D

E

F G

H

I J

K L

M N

P Q


Answers:

Example: The line marked on the coordinate grid below is one side of a square:

What are the possible coordinates of the corners of the square?

There are two possible places the square could be placed.

Students can visualize (or sketch) the

imaginary right triangle with the given line

as the hypotenuse.

By using it as a reference, students can then

find the location of the missing vertices of

the square.

LINKAGE: slope, rate of change

The missing coordinates could be at (6, 3)

and (1, 6) OR (-5, -4) and (0, -7).

5 units

5 units

a) b) c)


Example: Two corners of a square are shown in the coordinate plane below:

a) If the third corner is at (7, -1), where is the fourth

corner?

(5, 3)

b) If the third corner is at (-3, -1) where is the fourth

corner?

(-1, -5) It is possible to make a different square to those above

by placing the third and fourth points in two new

positions.

c) What are the coordinates which need to be plotted?

By looking at the given points as opposite vertices,

we can find the other corners at (4, 0) and (0, -2).

Example: Identify the coordinates of vertex D after quadrilateral DEFG is translated 7 units up:

(DOK 2)

A translation up will change the y-coordinate only and

move the figure up seven units. Therefore, D(3, -2) would

move to (3, 5). (b)

a) (3, -2)

b) (3, 5)

c) (5, 3)

d) (-4, -2)


Example: John rode his bike on Monday and again on Friday. His path for Monday is shown on

the graph below. Each unit represents 1 mile. (DOK 3)

On Friday, John rode 14 miles farther than he did on Monday. How could his path have changed

while remaining rectangular?

a) John could have ridden 14 miles farther north.

b) John could have ridden 3½ miles farther north and 7 miles farther east.

c) John could have ridden 7 miles farther east.

d) John could have ridden 1¾ miles farther north and 3½ miles farther east.

Students will need to determine the perimeter of the rectangle graphed to answer this

question.

Then… adding 14 would result in a larger perimeter of 26.

Students must remember that each change in a dimension

will result in twice as much change in the perimeter. Since

2 5 2 6

22

10 12

P

P

P

6 miles

5 miles


the perimeter must increase by 14, the only choice that will result in that amount of

change would be when John rides an additional 7 miles east. (d)

Surface Area

6.G.A.4 Represent three-dimensional figures using nets made up of rectangles and triangles,

and use the nets to find the surface area of these figures. Apply these techniques in the

context of solving real-world and mathematical problems.

Another way to look at three-dimensional figures is to look at a net. A net is an arrangement of

two-dimensional figures that can be folded to make three-dimensional figures. This will take the

student from two-dimensions to three-dimensions. Also have students start working the other

way: start with a three-dimensional solid, like a box, and see if they can draw what it would look

like if it was “unfolded” and laid flat. Students could even cut out their drawing and try to recreate

the solid.

The following websites will give you more resources:

http://www.mathisfun.com/platonic_solids.html printable nets for the platonic solids, shows

figures rotating (cube and tetrahedron only)

http://britton.disted.camosun.be.ca/jbpolytess.htm printable nets, tessellated in full color

http://www.senteacher.org/wk/3dshape.php printable nets for many different solids

In addition to drawing solid figures and working with nets, students are expected to create two-

dimensional drawings of three-dimensional figures and create three dimensional figures from a

two-dimensional drawing. For these notes and the creating of the practice test and test, we have

used Microsoft Word. Choose Insert Shapes then choose the cube in the Basic Shapes

section. You are then able to stack and build almost any 3-D shape of your choosing. Once your

figure is built you can “group” the figure to lock the shape. In class you can have students build 3-

D figures using wooden cubes, stacking cubes, interlocking cubes or Lego pieces to develop the

ability to see the top view, side view and front view.

Example: Given the following figure, identify (or draw) the top view, side view and front view.

From the top view, you would see…

From the front view, you would see…

From the side view, you would see…

http://www.mathisfun.com/platonic_solids.html

http://britton.disted.camosun.be.ca/jbpolytess.htm

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



From the top view, you would see…



Allow students to build and draw figures. As always, begin with very simple figures and allow

them to try more complex figures as they are able.


From the top view, you would see




Example: Given the top, side and front views, identify (or draw) the figure.

Answer:


Answer:


Answer:

Top View Side View Front View

Top View

Side View

Front View

Front View

Side View

Top View



Answer:

Surface Area

The surface area of a solid is the sum of the areas of all the surfaces that enclose that solid. To

find the surface area, draw a diagram of each surface as if the solid was cut apart and laid flat.

Label each part with the dimensions. Calculate the area for each surface. Find the total surface

area by adding the areas of all of the surfaces. If some of the surfaces are the same, you can save

time by calculating the area of one surface and multiplying by the number of identical surfaces.

Remind your students that “nets” are a way to break up these figures into surfaces for which we

can easily find the area.

For testing at the state level (grade 6) on surface area, only surface area nets made up of

from triangles and rectangles will be utilized.

Try to have students imagine the process of unfolding…

The following shows an example of the net of a triangular prism…

Top View

Front View

Side View


Example: Which of the following is NOT the net of a pyramid?

a)

b)

c)

d)

Answer: (b)

Example: Find the surface area of the prism shown.

All surfaces are squares.

Divide the prism into its parts.

Label the dimensions.

Bases Lateral Faces

Find the area of all the surfaces.

Bases Lateral Faces

Surface Area = Area of the top + bottom + front + back + side + side

Surface Area = 49 + 49 + 49 + 49 + 49 + 49

= 294

The surface area of the prism is 294 cm2.

7 cm

A = bh

A= 7 7

A = 49

A = bh

A= 7 7

A = 49

A = bh

A= 7 7

A = 49

A = bh

A= 7 7

A = 49

A = bh

A= 7 7

A = 49

A = bh

A= 7 7

A = 49

7 cm 7 cm 7 cm 7 cm 7 cm 7 cm

front back top bottom

side side


15 2

30

A bh

A

A

15 2

30

A bh

A

A

15 4

60

A bh

A

A

15 4

60

A bh

A

A

2 4

8

A bh

A

A

2 4

8

A bh

A

A

Since a cube has 6 congruent faces, a simpler method would look like

Surface Area = 6 the area of a face

Surface Area = 6B

Surface Area = 6bh

= 6 7 7

= 42 7

= 294

Again, the surface area of the prism is 294 cm2.

Example: Find the surface area of the prism shown.

All surfaces are rectangles.

Divide the prism into its parts.

Label the dimensions.

Bases Lateral Faces

Find the area of all the surfaces.

Bases Lateral Faces

Surface Area = Area of the top + bottom + front + back + side + side

Surface Area = 30 + 30 + 60 + 60 + 8 + 8

= 196

The surface area of the prism is 196 cm2.

top 2

15

bottom 2

15

front

15

4

back

15

4 sid

e

4

2

sid

e

4

2

2 cm

4 cm

15 cm

cm


Note: Since some of the faces were identical, we could multiply by 2 instead of adding the value

twice. That work would look like

Surface Area = 2(top or bottom) +2(front or back) +2(side)

Surface Area = 2(30) + 2(60) + 2(8)

= 60 + 120 + 16

= 196

Again, the surface area of the prism is 196 cm2.

Example: Find the surface area of the triangular prism.

If we break our triangular prism down into a net, we get this:

In a triangular prism there are five faces, two

triangles and three rectangles.

The total surface area would be the sum of all

the areas…

36.45 72.9 58.32 18 18

203.67S

SA

A

The surface area is 203.67 cm2.

1

9.0 4

1

8

2

1

2b

A

A

h


Example: Find the surface area of the trapezoidal prism.

A net of this solid would look something like this…

Filling in measurements…

With the both bases…

The total surface area would be the sum of all the areas…

400 780 260 360 342 34

24

2

84SA

SA

The surface area is 2484 m2.

20 X 20

= 400

39 X 20

= 780

13 X 20

= 260

18 X 20

= 360

118 39 12

2

157 1

34

22

2A

A

A


Example: Find the surface area of the isosceles trapezoidal prism.


15 15 15 34 32 32

143SA

SA

The surface area is 143 in2.

Example: Find the surface area of the triangular prism.


312

96 96 96 12 12SA

SA

The surface area is 312 cm2.

15 11 4

2

116

3

4

2

2

A

A

A

1

8 3

2

12

2

1bh

A

A

8

3

8

12

8

3

5 5 5

5

5 5

11 4


Example: Find the surface area of the rectangular pyramid.

a) 206 m2

b) 312 m2

c) 302 m2

d) 216 m2


312

96 48 48 60 60SA

SA

The surface area is 312 m2.

Example: The surface area of the composite solid of the figure below is

a) 5000 cm

2

b) 4950 cm2

c) 4550 cm2

d) 4450 cm2

12

8 12

10

1

2

4

2

8 12

8

1

A

h

A

b

1

12 10

1

0

2

6

2b

A

A

h


The net for this solid consists of six rectangles and two oddly shaped bases.


800 200 200 1000 600 1200 60 475 475

5010

S

A

A

S

The surface area is 5010 cm2.

Optional Extension:

The formula for Total Surface Area of a rectangular prism is given as:

In addition, a discussion about the difference between Total Surface Area and Lateral

Area can be introduced. Lateral Area is defined to be the surface area of a three-

dimensional object minus the area(s) of the base(s). We can call the faces included in the

Lateral Area the lateral faces.

Can you find the Lateral Area of the solid in the previous example?

20

20

40

5 5 25 15 30

5

5

40 X 20

= 800 40 X 25

= 1000

40 X 15

= 600 40 X 30

= 1200

40 X

5 =

200

40 X

5 =

200

5 X 5 = 25

30 X 15 = 450

Decomposed

Area of base =

450 + 25 = 475


Notice the absence of the two bases in the following diagram.

The sum of the areas of the six rectangles would be the Lateral Surface Area.

It can be found by adding the areas of each lateral face…

800 200 200 1000 600 1200 60

4060

L

LA

A

OR

by taking the Total Surface Area and subtracting the areas of the two bases…

5010 (475 475)

4060

A

A

L

L

The Lateral Area is 4060 cm2.

Example: An area to be planted with grass seed measures 50 feet by 75 feet. Before planting, a

3-inch layer of loam is spread on the area.

Part A: How many cubic feet of loam is needed?

Part B: A truck delivers loam in cubic yards. The landscaper divides the cubic feet of loam is

needed by 9 to find the cubic yards that will be needed. Will this calculation produce the

correct results? Explain your answer.

Part C: How many cubic yards of loam will need to be delivered?

Answer:

Part A: Since the units are not the same they must be converted. We can review the ratio

conversions in this process.

The loam would create a rectangular prism on top of the garden that is 50 75 3ft ft in ,

either both 50 ft and 75 ft must be converted or 3 inches must be converted.

Let’s see what happens both ways…

50 12600

1 1

ft inin

ft

AND

75 12900

1 1

ft inin

ft

So the new dimensions are… 600 900 3in in in

20

40

5 5 25 15 30


Using those dimensions we can find the volume in cubic inches.

The volume is 1,620,000 in3 which must be

converted back into cubic feet.

MISCONCEPTION: Since each foot contains 12 inches…students want to divide by 12 to

obtain the answer. WRONG!!

They must actually

divide by 1,728!!

That’s a BIG

difference!

1,620,000 1 9 .28 57 37 937.5 ft3 of loam will be needed.

OR

3 1 3 1

1 12 12 4

in ftft ft

in

So the new dimensions are…1

50 754

ft ft in .

The volume is 937.5 ft3.

Notice, in this case, no further conversions are necessary. Nice convenience, but we did have to

work with fractions.

Part B: Again, watch for that misconception when converting. The correct conversion is shown

below.

So… NO, the landscaper is wrong! He will order too much, because a cubic yard is 27 cubic

feet he needs to divide by 27.

Part C: 3 3

3 33

3

937.5 1 937.534.72

1 27 2735

ft ydyd yd

fd

ty

35 cubic yards of loam will be delivered.

600 900

1,620,0

3

00V

V

1

1

1 1

V

V

1 ft. 12 in.

12 12 12

1,728V

V

9

150

37.5

754

V

V

1

1

1 1

V

V

1 yd 3 ft

3 3

27

3

V

V

1 cubic foot 1,728 cubic inches

1 cubic yard

yard

1 cubic foot

27 cubic feet

yard

1 cubic foot


Sample SBAC Questions

Standard: 6.G.1, 6.G.3 DOK: 2 Item Type: TE Difficulty: M



Standard: 6.G.2, 6.NS.3 DOK: 2 Item Type: ER Difficulty: M



Standard: 6.G.4 DOK: 2 Item Type: TE Difficulty: M



Sample Explorations Questions

Correct Answer: A

Correct Answer: C

Correct Answer: B


Correct Answer: A

Correct Answer: A

Correct Answer: A


Correct Answer: A

Correct Answer: B

Correct Answer: B


Correct Answer: C

Correct Answer: B


Correct Answer: B

Correct Answer: B

Documents

Math 6: Geometry Notes 2-Dimensional and 3-Dimensional …rpdp.net/admin/images/uploads/resource_6228.pdf · 2019-04-21 · Math 6 Notes Geometry: 2-Dimensional and 3-Dimensional