Math 6 Unit 9 Notes: Measurement and Geometry, Area/Volume
Perimeter
Objectives: (5.5) The student will model formulas to find the perimeter, circumference and
area of plane figures.
(5.6) The student will apply formulas to find the perimeter, circumference and
area of plane figures.
Previously in Unit 8 notes, we worked on the conceptual understanding, modeling of and
deriving the formulas for perimeter, circumference and area of plane figures. Now we are
moving into the application of these formulas.
The perimeter of a polygon is the sum of the lengths of the segments that make up the sides of
the polygon.
2m
Example: Find the perimeter of the regular pentagon.
Since the pentagon is regular, we know all five sides have a measurement of 2 meters.
So we simply multiply 5 2 for an answer of 10 meters for the perimeter.
Example: Find the perimeter for a rectangle with length 7 feet and width 2 feet.
Since opposite sides of a rectangle are equal in length, P
2 7
2 2 or 18 feet.
Example: The perimeter of a square is 64 inches. Find the length of each side.
P= 4s
64=4s
16=s
Each side of the square is 16 inches.
Example: Find the length of the missing side if you know the perimeter is 67 meters.
k
Perimeter is equal to the sum of the sides.
67
13 15 13 k
67
k
26
k
13 m
13 m
41
15 m
The length of the missing side is 26 meters.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 1 of 27
6 mm
Example: Find the perimeter for the polygon.
4 mm
1. Students must determine the
dimensions of the missing side lengths.
4 mm
Since this is a rectangle (opposite sides
congruent) and the bottom length is 10 mm
and the top piece is 6 mm, then 10 6=4.
The left side of the rectangle is 8 mm
and on the right we have 4 mm, so the far right
section must be 8 4=4.
2. Then compute the perimeter by finding
the sum of the edge lengths.
or
P = 6+4+4+4+10+8
P= 36
4 mm
Find the missing dimensions.
8 mm
10 mm
since bottom = 10, top pieces =10
since left =8, right pieces=8
SO, P=10+10+8+8
P= 36
The perimeter is 36 mm.
Example: Find the perimeter for the polygon.
P = 4+3+6+2+6+3+4+8
P = 36
8-3-3=2
The perimeter is 36 units.
Circles: Circumference
A circle is defined as all points in a plane that are equal distance (called the radius) from a fixed
point (called the center of the circle). The distance across the circle, through the center, is called
the diameter. Therefore, a diameter is twice the length of the radius, or d 2r .
centerpoint
Math 6 Notes
diameter
Unit 09: Measurement and Geometry
Revised 2012 CCSS
radius
Page 2 of 27
We called the distance around a polygon the perimeter. The distance around a circle is called the
circumference. There is a special relationship between the circumference and the diameter of a
circle. Let’s get a visual to approximate that relationship. Take a can with 3 tennis balls in it.
Wrap a string around the can to approximate the circumference of a tennis ball. Then compare
that measurement with the height of the can (which represents three diameters). You will
discover that the circumference of the can is greater than the three diameters (height of the can).
There are many labs that will help students understand the concept of pi, . Most involve
measuring the distance around the outside of a variety of different size cans, measuring the
diameter of each can and dividing to show that the relationship is “3 and something”. You may
want to use a table like the following to organize student work:
Can #
Circumference
Diameter
Circumference
Diameter =
Once students record the measurements for a variety of cans, students should find the quotient of
the circumference and the diameter is close to 3.14. Some quotients are not exact, or are “off”,
due to measurement errors or lack of precision.
So initially get students to comprehend the circumference
diameter x 3.
This should help convince students that this ratio will be the same for every circle.
C
π or C πd . Then, since d 2r , we can also write C 2πr .
We can then introduce that
d
Please note that π is an irrational number (never ends or repeats). Mathematicians use to
represent the exact value of the circumference/diameter ratio.
The formulas for the circumference of a circle are C
d and C
2 r.
Example: A circle has a radius of 4 m. Estimate the circumference of the circle. (Use 3 to
approximate pi.)
Using the formula:
C
2 r
C
234
C
24
The circumference is about 24 meters.
Note: Many standardized tests (including the CRT and the district common exams) ask students
to leave their answers in terms of π. Be sure to practice this!
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 3 of 27
Example: If a circle has a radius of 5 feet, find its circumference. Do not use an approximation
for π.
Using the formula:
C
2πr
C
2 π 5
C
10π
The circumference is 10π feet.
Example: If a circle has a diameter of 35 inches, find its circumference. Leave your answer in
terms of π.
C
Using the formula:
d
C
35
C
The circumference is 35π inches.
35
Example: A circle has a diameter of 24 m. Using π 3.14 , find the circumference. Round your
answer to the nearest whole number.
Using the formula:
C
πd
C
(3.14) 24
C
75.36
The circumference is about 75 meters.
Example: If a circle has a diameter of 4 m, what is the circumference? Use 3.14 to approximate
π. State your answer to the nearest 0.1 meter.
Using the formula:
CRT Example:
C
πd
C
(3.14)(4)
C
12.56
The circumference is about 12.6 meters.
Solution:
The circumference of the original young tree would be:
C
d
3.14 20
62.8
The circumference of the older tree would be:
C
d
3.14 23
72.22
The difference between the two circumferences is:
72.22
-62.8
9.42
Answer:
Math 6 Notes
B
9.42 cm
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 4 of 27
Area of Plane Figures
Area of Rectangles, Squares and Rhombi
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. That’s the size of the
closet.
2 ft.
6 ft.
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.
2 ft.
6 ft.
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. More formally, area is defined as the space inside a figure or the amount of surface a
figure covers. 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 A bh .
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 A lw . So now we have 2 formulas for the areas of a rectangle
that can be used interchangeably. The answer in an area problem is always given in square units
because we are determining how many squares fit inside the region. Of course you will show a
variety of rectangles to your students and practice identifying the base and height of those
various rectangles.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 5 of 27
Example: Find the area of a rectangle with the dimensions 3 m by 2 m.
A
lw
A
3 2
A
6
The area of the rectangle is 6 m2.
Example: The area of a rectangle is 16 square inches. If the height is 8 inches, find the base.
Solution:
A = bh
16 = b 8
2=b
The base is 2 inches.
Example: The area of the rectangle is 24 square centimeters. Find all possible whole number
dimensions for the length and width.
Solution:
Area Length
(in cm)
24
1
24
2
24
3
24
4
24
6
24
8
24
12
24
24
Width
(in cm)
24
12
8
6
4
3
2
1
9 ft.
Example: Find the area of the rectangle.
2 yd.
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.
Since 1 yard = 3 feet we start with
yards
feet
1
3
x
9
9
3x
3
x
We now have a rectangle with dimensions 3 yd. by 2 yd.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 6 of 27
A
A
A
lw
3 2
6
The area of our rectangle is 6 square yards.
height
If we were to have a square whose sides measure 5 inches, we could find the area of the square
by putting it on a grid and counting the squares as shown below.
Again we could count the boxes to find the area of the square
( 25units 2 ), but more easily we could multiply the base height or
the length the width ( 5 x5 25 units 2 ) . So again, A=bh or A lw .
Students may also see that since the base and height or length and
width of a square are congruent, they may choose to use A s 2
( 5 x5 25 units 2 ).
base
Counting the squares is a viable method but eventually students
will begin to see the area can be computed more efficiently. If they
multiply the base times the height, then again A bh or some will
see the area as the square of one side, so A s 2 .
In this example, A bh 5 5 25 units 2 or
A s 2 52 5 5 25 units 2 .
height
height
Next, if we begin with a 6 x 6 square and cut from one corner to the other side (as show in light
blue) and translate that triangle to the right, it forms a new quadrilateral called a rhombus (plural
they are called rhombi). Since the area of the original square was 6 x 6 or 36 square units, then
the rhombus has the same area, since the parts were just rearranged. So we know the rhombus
has an area of 36 square units. Again we find that the Area of the rhombus = bh.
base
base
Be sure to practice with a number of different rhombi so your students are comfortable with
identifying the height versus the width of the figures.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 7 of 27
Area of Parallelograms, Triangles and Trapezoids
height
height
If I were to cut one corner of a rectangle and place it on the other side, I would have the
following:
base
base
We now have 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.
So we have the Area of a parallelogram = bh.
h
b
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, A bh
A
3 6
A 18
The area of the parallelogram is 18 m2 .
NOTE: If students are having difficulty with correctly labeling areas with square units
considering having them write the units within the problem. For example, instead of
A
bh
A
3 6
A 18
try
A
bh
A
3 meters 6 meters
A 18 square meters
Say aloud 3 6 = 18
and meters meters
= square meters
18 square meters
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 8 of 27
We have established that the area of a parallelogram is A bh . Let’s see how that helps us to
understand the area formula for a triangle and trapezoid.
For this parallelogram, its base is 4 units
and its height is 3 units. Therefore, the
area is 4 3 12 units 2 .
h
base
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,
1
1
A
bh, or
4 3 6 units 2
2
2
h
base
For this parallelogram, its base is 8 units
and its height is 2 units. Therefore, the
area is 8 2 16 units 2 .
h
base
b1
b2
h
b2
base
b1
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
b1 and b2 . For a trapezoid:
1 1
A A (b1 ( bb1 2 ) hb2 )h
2 2
Triangles
As demonstrated above, one way to introduce the area formula of a triangle is to begin with a
rectangle, square, rhombus and/or a parallelogram and show that cutting the figure using a
diagonal produces 2 congruent triangles. This aids students in remembering the Area formula of
1
bh
a Triangle is A
.
bh or A
2
2
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 9 of 27
Example: Find the area of each of the following triangles.
b
..
a.
11 cm
10 in
8 in
13 cm
6 cm
c.
8 mm
25 mm
16 mm
12 mm
17 cm
6 in
Solutions:
A
1
bh
2
1
(17)(6)
2
(17)(3)
A
51
A
a.
A
A
1
bh
2
1
(6)(8)
2
3(8)
A
24
A
b.
51 square cm
A
24 squares inches
A
1
bh
2
1
(12)(8)
2
6(8)
A
48
A
c.
A
48 square mm
Important things here:
In each example, students must determine which dimensions given are the base and
height.
In example a, the Commutative Property of multiplication was employed to take half of
the even number to make the arithmetic easy.
In example c, the height must be given outside the triangle. Why?
Example: A triangular piece of fabric has an area of 54 square inches. The height of the triangle
is 6 inches. What is the length of the triangle’s base?
Beginning with the formula and substituting in the values we know, we get:
1
A
bh
2
1
The triangle’s base is 18 inches.
54
(b )(6)
2
54 3(b )
18
b
Reflection:
When can you use two side lengths to find the area of a triangle? In this situation, does it matter
which side is the base and which side is the height?
Solution: When you have a right triangle you use two side lengths to find the area of a
triangle. It doesn’t matter which side is the base or height.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 10 of 27
CRT Example:
A
Solution:
A
1
bh
2
1
12 8
2
Trapezoids
As demonstrated on page 9, the formula used to find the Area of a trapeziod is
1
A
b1 b2 h.
2
We must be sure students see this formula in several ways. Be sure to teach the other forms
b b
Area of a trapeziod is A ( 1 2 )h and A ( Average of the bases ) height .
2
4 ft
Example:
a.
4 ft
20 mm
b.
c.
8 mm
6 mm
6 ft
5 cm
2 cm
24 mm
8 cm
Solutions:
a.
A
( Average of the bases ) height
A
54
A
20
b.
A
1
(b1 b2 ) h
2
1
(5 2)(8)
2
1
(7)(8)
2
4(7)
A 132
A
28
132 mm 2
28 cm 2
A
A
A
A
20 ft 2
Math 6 Notes
b1 b2
)h
2
20 24
(6)
2
44
(6)
2
22(6)
(
Unit 09: Measurement and Geometry
Revised 2012 CCSS
c.
A
A
A
Page 11 of 27
Notice:
In example a, it was easy to mentally find the average of 6 and 4 (the bases) thus the use
of the formula A= (Average of the bases) height.
In example c, students will need to correctly identify the bases to solve this problem. Be
sure to give students exposure to this orientation of the trapezoid.
Example: A trapezoid has an area of 200 cm 2 and bases of 15 cm and 25 cm. Find the height.
Solution:
1
(b1 b2 ) h
2
1
200
(15 25) h
2
1
200
(40) h
2
200 20 h
A
10
The height is 10 cm.
h
Area of Circles
You can demonstrate the formula for finding the area of a circle. First, draw a circle; cut it out.
Fold it in half; fold in half again. Fold in half two more times, creating 16 wedges when you
unfold the circle. Cut along these folds.
Rearrange the wedges, alternating the pieces tip up and down (as shown), to look like a
parallelogram.
radius (r)
This is ½ of the distance around the
circle or ½ of C. We know that
C
1
C
2
1
C
2
Math 6 Notes
2πr , so
1
2πr
2
πr
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 12 of 27
The more wedges we cut, the closer it would approach the shape of a parallelogram. No area has
been lost (or gained). Our “parallelogram” has a base of πr and a height of r. We know from
our previous discussion that the area of a parallelogram is bh. So we now have the area of a
circle:
A bh
radius (r)
A (πr )( r )
A
πr
πr 2
Example: A circle has a radius of 4 m. Estimate the area of the circle. (Use 3 to approximate pi.)
Using the formula:
A
r2
A
3 42
A
3 16
A
48
The area of the circle is approximately 48 m 2
Example: A student estimated the area of the circle to the right as shown. Is the solution
correct? Explain your thinking.
A
r2
Solution:
8 mm
A 3 82
A
3 64
A 192 mm 2
Example: Find the area of the circle in terms of pi, if the diameter is 20 feet.
Students must note:
Using the formula:
If diameter is 20 feet, then the radius is 10 feet.
A
r2
A
10 2
A
100
The area of the circle is 100
feet 2 .
A 100
Example: Find the area of the circle if the diameter is 10 inches. Leave your answer in terms of
π.
Using the formula:
A
πr 2
A
π 5
2
Remember if the diameter is 10
inches, the radius must be 5 inches.
A 25π
The area of the circle is 25π square inches.
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 13 of 27
Many standardized tests (including the CRT and the district common exams) ask students to
leave their answers in terms of π. Be sure to practice this!
Example: Find the area of the circle to the nearest square meter if the radius of the circle is
12 m. Use
3.14 .
Using the formula:
A
πr 2
A
(3.14)(12) 2
A
(3.14)144
A
452.16
The area of the circle is about 452 square meters or 452 m 2 .
Example: Find the area of the circle if the diameter of the circle is 200 m. Use
Students must note:
If diameter is 200 m, then the radius is 100 m.
A
Using the formula:
3.14 .
r2
A
3.14 100 2
A
3.14 10, 000
A
31, 400
The area of the circle is 31,400 square meters or 31,400 m 2 .
Example: Find the area of the circle if the radius of the circle is 7 m. Use
A
A
Using the formula:
A
A
22
.
7
r2
22 2
7
7
22 7 7
7 1 1
22 7
The area of the circle is 154 m 2 .
A 154
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 14 of 27
Example: Find the area of the circle if the diameter is 21 inches. Use
A
A
r2
22
7
11
22
A
7
1
Using the formula:
A
A
A
22
.
7
21
2
2
3
21 21
2 2
1
The area of the circle is 346
1 2
in .
2
11 3 21
2
693
2
1
346
2
Reflection - Check for Understanding
For which plane figure(s) does the formula A bh work?
For which plane figure(s) does the formula A lw work?
For which plane figure(s) does the formula A
For which plane figure(s) does the formula A
For which plane figure(s) does the formula A
For which plane figure(s) does the formula A
For which plane figure(s) does the formula A
1
bh work?
2
s 2 work?
r 2 work?
( Average of the bases ) height work?
1
(b1 b2 ) h work?
2
Example: When a tree was planted, its diameter measured 6 cm. Five years later, the diameter
of the tree measured 10 cm. Find the difference between the two areas. (Use =3.14)
Solution:
The area of the original young tree would be:
A
r2
3.14 32
28.26
The area of the older tree would be:
A
r2
3.14 5 2
3.14 25
78.5
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 15 of 27
The difference between the two areas is:
78.50
- 28.26
50.24
The difference in the areas is 50.24 cm 2 .
Answer:
Area of Composite Figures
Students should also practice finding the area of irregular figures by composing and
decomposing into shapes they know.
6 cm
Example: Find the area of the polygon.
8 cm
5 cm
10 cm
Solutions: Method 1(Whole to Part) Composing
6 cm
8 cm
1. Create a rectangle using the outside borders of the
original figure. (Composing)
2. Find the area of the large “new” rectangle.(Whole)
3. Subtract the negative (white) space. (Part)
5 cm
10 cm
3
A=
8
4
10
A 10 8
34
A
80 12
A
68
Area = 68cm 2
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 16 of 27
Method 2 (Part to Whole) Decomposing
6 cm
8 cm
5 cm
1. Break the polygon into smaller figures (rectangles,
squares, triangles, etc. that you know). (Decomposing)
2. Find the area of each piece.(Parts)
3. Add the areas of the pieces. (Whole)
10 cm
A=
+
8
4
6
A
68 45
A
48
A
68
5
20
Area = 68 cm 2
Example: Find the area of the polygon.
1
1
1
1
1
1
1
1
Solutions: Method 1(Whole to Part)
1
1
1
1
5
1. Create a rectangle using the outside borders of the
original figure. (Composing)
2. Find the area of the large “new” rectangle.(Whole)
3. Subtract the negative (white) spaces. (Parts)
1
1
1
1
6
Math 6 Notes
NOTE: Students will need to compute
outer side length and width (shown in
red).
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 17 of 27
5
A=
4(
6
A
6 5
A
30
A
26
1
1 )
4 1 1
Area = 26 square units
4
Method 2
(Part to Whole)
Method 3
(Part to Whole)
1. Break the polygon into smaller figures (rectangles,
squares, triangles, etc. that you know). (Decomposing)
2. Find the area of each piece.(Parts)
3. Add the areas of the pieces. (Whole)
1
1
1
1
1
1
1
1
5
1
1
1
1
1
1
1
1
66
3 +
A=
6
4
1 +
1
4
5
A=
4
A=
36
+
14
+
14
A=
18
+
4
+
4
A=
26
A= 45
A=
+ 3 + 3
1
1
+ 31 + 31
20 + 3 + 3
A = 26
26 units 2
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 18 of 27
5m
Example: Find the area of the polygon.
20 m
14 m
10 m
Solution:
Method 1
6
A=
Method 2
5
5
6
20
A
1
65
2
200 15 15
A
170
10 20
14
+
1
65
2
6
10
10
10
A
A=
A
1
10 6
2
140 30
A
170
A
10 14
170 m 2
170 m 2
NOTE: Students will need to compute the outer side length and width
(shown in red).
Example: The dimensions of a church window are shown below.
Find the area of the window to the nearest square foot.
8 feet
A=
+
10 feet
1 2
r
2
1
A 10 8
(3.14)(5) 2
2
A 80 1.57(25)
A
bh
A 80
We are given the diameter, so the radius would be
half of the 10 feet or 5 feet.
39.25
A 119.25
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 19 of 27
The area of the church window is about 119 square feet to the nearest foot.
Comparing Perimeter and Area
In this section students are given polygons that are either enlarged (dimensions doubled or
tripled) or reduced (dimensions halved) to create similar figures, and students are asked to
compare the perimeters and/or areas. Many students will need to build or draw these figures to
visually and mentally determine “what happens when”. the dimensions are…
When dimensions are doubled:
Original
Figure
Example 1:
Dimensions
Doubled
1 cm
2 cm
1 cm
2 cm
Perimeter
P
2(l
P
w)
P
2(l
2(1 1)
P
2(2 2)
P
2(2)
P
2(4)
P
4
P
8
4 cm
w)
8 cm
Area
A
l w
A
l w
A 11
A
22
A 1
A
4
1 cm
Example 2:
2
4 cm 2
Original
Figure
Dimensions
Doubled
4 units
8 units
2 units
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 20 of 27
4 units
Perimeter
P
2(l
w)
P
2(l
P
2(2 4)
P
2(4 8)
P
2(6)
P
2(12)
P
12
P
24
12 units
w)
24 units
Area
A
l w
A
l w
A
24
A
48
A
8
A
32
8 units 2
32 units 2
Example 3:
Dimensions
Doubled
Original
Figure
2 units
4 units
2 units
4 units
Perimeter
P
2(l
w)
P
2(l
P
2(2 2)
P
2(4 4)
P
2(4)
P
2(8)
P
8
P
16
8 units
w)
16 units
Area
A
l w
A
l w
A
22
A
44
A
4
A 16
4 units 2
Math 6 Notes
16 units 2
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 21 of 27
Charting the perimeters from Examples 1 -3 when the dimensions of a figure are DOUBLED we
find the following:
Perimeter of original Perimeter of figure doubled
Example 1
4 cm
8 cm
Example 2
12 cm
24 cm
Example 3
8 cm
16 cm
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are doubled, the perimeter is doubled.
Charting the area from Examples 1 -3 when the dimensions of a figure are DOUBLED we find
the following:
Area of original Area of figure doubled
Example 1
1 cm 2
4 cm 2
Example 2
8 cm 2
32 cm 2
Example 3
4 cm 2
16 cm 2
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are doubled, the area is four times or 22 .
When dimensions are tripled:
Example 4:
Dimensions
Tripled
Original
Figure
3 cm
1 cm
1 cm
3 cm
Perimeter
P
2(l
P
P
2(l
2(1 1)
P
2(3 3)
P
2(2)
P
2(6)
P
4
P
12
4 units
Math 6 Notes
w)
w)
12 units
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 22 of 27
Area
A
l w
A
l w
A 11
A
33
A
9
A 1
1 units
2
9 units 2
Example 5:
Dimensions
Tripled
Original
Figure
4 units
12 units
2 units
6 units
Perimeter
P
2(l
w)
P
2(l
w)
P
2(2 4)
P
2(6 12)
P
2(6)
P
2(18)
P
12
P
36
36 units
12 units
Area
A
l w
A
l w
A
24
A
6 12
A
8
A
72
8 units
2
72 units 2
Example 6:
Dimensions
Tripled
Original
Figure
1 in
3 in
5 in
15 in
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 23 of 27
Perimeter
P
2(l
w)
P
2(l
w)
P
2(5 1)
P
2(15 3)
P
2(6)
P
2(18)
P
12
P
36
36 units
12 units
Area
A
l w
A
A
51
A 15 3
A
5
A
5 units 2
l w
45
45 units 2
Charting the perimeters from Examples 4-6 when the dimensions of a figure are TRIPLED we
find the following:
Perimeter of original Perimeter of figure tripled
Example 4
4 units
12 units
Example 5
12 units
36 units
Example 6
12 units
36 units
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are tripled, the perimeter is tripled.
Charting the area from Examples 4-6 when the dimensions of a figure are TRIPLED we find the
following:
Area of original Area of figure tripled
Example 4
1 units 2
9 units 2
Example 5
8 units 2
72 units 2
Example 6
5 units 2
45 units 2
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are cut in tripled, the area is 9 times or 3 2 .
When dimensions are halved:
Original
Figure
Example 7:
Halved
Figure
1 cm
2 cm
1 cm
2 cm
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 24 of 27
Perimeter
P
2(l
w)
P
2(l
P
2(2 2)
P
2(1 1)
P
2(4)
P
2(2)
P
24
P
4
8 cm
4 cm
A
l w
A
A
22
A 11
A
4
A 1
4 cm 2
1 cm 2
w)
Area
Original
Figure
l w
Halved
Figure
Example 8:
2 units
4 units
1 unit
2 units
Perimeter
P
2(l
w)
P
2(l
P
2(2 4)
P
2(1 2)
P
2(6)
P
2(3)
P
12
P
6
12 cm
6 cm
A
l w
A
A
24
A 12
A
8
A
w)
Area
8 cm 2
Math 6 Notes
l w
2
2 cm 2
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 25 of 27
Original
Figure
Halved
Figure
Example 9:
3 units
6 units
2 units
4 units
Perimeter
P
2(l
w)
P
2(l
P
2(4 6)
P
2(2 3)
P
2(10)
P
2(5)
P
20
P
10
20 cm
10 cm
A
l w
A
l w
A
46
A
23
A
24
A
6
w)
Area
24 cm 2
6 cm 2
Charting the perimeters from Examples 7-9 when a figure is HALVED we find the following:
Perimeter of original Perimeter of figure halved
Example 7
8 units
4 units
Example 8
12 units
6 units
Example 9
20 units
10 units
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are cut in half, the perimeter is cut in half.
Charting the area from Examples 7-9 when a figure is HALVED we find the following:
Area of original Area of figure halved
Example 7
4 units 2
1 units 2
Example 8
8 units 2
2 units 2
Example 9
24 units 2
6 units 2
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
Page 26 of 27
Looking at these example results, the pattern that occurs is that when the dimensions of a figure
are cut in half, the area is cut to one- fourth or
1
2
2
.
In summary, when the dimensions of a rectangle are: doubled, the perimeter is doubled.
tripled, the perimeter is tripled.
halved, the perimeter is halved.
In summary, when the dimensions of a figure are: doubled, the area is 22 or four times as great.
tripled, the area is 3 2 or nine times as great.
1
halved, the area is
2
Math 6 Notes
Unit 09: Measurement and Geometry
Revised 2012 CCSS
2
or one fourth as great.
Page 27 of 27