1.11 length and perimeter w

Length

Length and Perimeter

http://www.lahc.edu/math/frankma.htm


Length
In the beginning, to measure and record a length, we gauged
and matched the measurement to a physical item such as a
stick or a piece of rope (we all have dipped sticks in pools of
water to measure their depths.)


Length
After units such as feet and meters were established, we were
able to record lengths with numbers instead of physical objects.


Length
For example, to find the distance between two locations L and R
with L at 4 and R at 16 as shown below,
0

2

4

16

L

R


Length
we subtract: 16 – 4 = 12, so that they are 12 units apart.
0

2

4

16

L

R


Length
2

4

16

L

0

R

In general, to find the distance between two locations L and R
on a ruler, we subtract: R – L (i.e. Right Point – Left Point).
L

R


Length
2

4

16

L

0

R

In general, to find the distance between two locations L and R
on a ruler, we subtract: R – L (i.e. Right Point – Left Point).
L

R

The length between L and R is R – L

Example A. a. Find the mileage-markers for points A and B
and the distance between A and B below.

0

B

A

60 (miles)

The 60-mile segment is divided into 12 pieces, so each
subdivision is 60/12 = 5 miles.
0

B

A

60 (miles)

subdivision is 60/12 = 5 miles. Hence, B is at 25 and A is at 45.
0

B

A

60 (miles)

0

B

A

25

45

60 (miles)

The distance between A and B is 45 – 25 = 20 miles.
0

B

A

25

45

60 (miles)

B

A

25

0

45

60 (miles)

b. The following is a straight road and the corresponding
positions of towns A, B, and C, marked along the road.
A

B

C

We drove from A to B and found out they are 56 miles apart.
How far is A to C? How many miles more are there to reach C?

B

A

25

0

45

60 (miles)

A

B

C

There are 7 subdivisions from A to B which covers 56 miles,
hence each sub-divider is 56/7 = 8 miles.

B

A

25

0

45

60 (miles)

A

B

C

hence each sub-divider is 56/7 = 8 miles. There are 12
subdivisions from A to C, i.e. so they are 8 x 12 = 96 miles.

B

A

25

0

45

60 (miles)

A

B

C

hence each sub-divider is 56/7 = 8 miles. There are 12
subdivisions from A to C, i.e. so they are 8 x 12 = 96 miles.
There are 96 – 56 = 40 more miles to reach C.

If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.

we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.

we obtain a loop.
The loop forms a
perimeter or border
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.

we obtain a loop.
The loop forms a
perimeter or border
All the areas above are enclosed by the same rope,
so they have equal perimeters.

we obtain a loop.
The loop forms a
perimeter or border
A plane-shape is a polygon if it is formed by straight lines.

we obtain a loop.
The loop forms a
perimeter or border
Following shapes are polygons:

we obtain a loop.
The loop forms a
perimeter or border
Following shapes are polygons:
These are not polygons:

Three sided polygons
are triangles.

are triangles.
Triangles with three equal sides
are call equilateral triangles.

are triangles.

s
s
s

An equilateral triangle

are triangles.

s
s
s

Triangles are different from other polygons because if all three
sides of a triangle are known then the shape of the triangle is
determined.

are triangles.

s
s
s

determined. This is not so if a polygon has four or more sides.

are triangles.

s
s
s

For example, there is only
one triangular shape with
all three sides equal to 1.
1

1
1

s

are triangles.

s
s

For example, there is only Four-sided polygons with sides
one triangular shape with equal of 1 may be squashed into
various shapes.
1

1
1

1

1

1

1

s

are triangles.

s
s

various shapes.
1

1

1

1

1
1
1
Because of this, we say that “triangles are rigid”,

s

are triangles.

s
s

various shapes.
1

1

1

1

1
1
1
Because of this, we say that “triangles are rigid”,
and that in general “four or more sided polygons are not rigid”.

If the sides of a triangle are labeled as a, b, and c, then
a + b + c = P, the perimeter.
b

c

a
P=a+b+c

a + b + c = P, the perimeter. The perimeter of an equilateral
triangle is P = s + s + s = 3s.
b

c

a
P=a+b+c

b
Rectangles are 4-sided polygons where
the sides are joint at a right angle
c
as shown.
a
P=a+b+c

b
c
as shown. A square
a
s
P=a+b+c
s

s
s

b
c
as shown. A square
a
s
P=a+b+c
s

s

s
A square is a rectangle with four equal sides.
The perimeter of a squares is P = s + s + s + s = 4s

b
c
as shown. A square
a
s
P=a+b+c
s

s

s
If we know two adjacent sides of a
rectangle, we know the entire rectangle
because the opposites sides are identical.

b
c
as shown. A square
a
s
P=a+b+c
s

s

s
If we know two adjacent sides of a
rectangle, we know the entire rectangle
because the opposites sides are identical.

However the names of the two sides is a source of confusion.

By the dictionary, “length” is the longest
side and “width” is the horizontal side.

This causes conflicts in labeling rectangles
whose width is the longer side.

length? width?

length? width?
Hence we will use the words “height” for the vertical side,
and “width” for the horizontal side instead.
width (w)
h

height (h)
w

length? width?
width (w)
The perimeter of a rectangle is
P = 2h + 2w (= h + h + w + w)
h
height (h)
w

length? width?
width (w)
P = 2h + 2w (= h + h + w + w)
h
height (h)
Example B. a. We drove in a loop as shown.
How many miles did we travel?
w
Assume it’s an equilateral triangle on top.
5 mi
3 mi

length? width?
width (w)
P = 2h + 2w (= h + h + w + w)
h
height (h)
w
There are three 3-mile sections and two 5-mile
sections.

5 mi
3 mi

length? width?
width (w)
P = 2h + 2w (= h + h + w + w)
h
height (h)
w
5 mi
3 mi

sections. Hence one round trip P is
P=3+3+3+5+5
= 3(3)+ 2(5)

length? width?
width (w)
P = 2h + 2w (= h + h + w + w)
h
height (h)
w
5 mi
3 mi

sections. Hence one round trip P is
P=3+3+3+5+5
= 3(3)+ 2(5) = 9 + 10 = 19 miles.

b. We want to rope off a 50-meter by 70-meter rectangular
area and also rope off sections of area as shown.
How many meters of rope do we need?
70 m

50 m

70 m
We have three heights where each
requires 50 meters of rope,

50 m

70 m
and three widths where each
50 m
requires 70 meters of rope.

70 m
50 m
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.

70 m
50 m
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
c. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft

70 m
50 m
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
2 ft
The perimeter of the step-shape is the
same as the perimeter of the rectangle
that boxes it in as shown.

70 m
50 m
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
2 ft
that boxes it in as shown. There are 3 steps
going up, and 5 steps going across.

70 m
50 m
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
2 ft
that boxes it in as shown. There are 3 steps
going up, and 5 steps going across. So the height is 6 ft,
the width is 10 ft, and the perimeter P = 2(6) +2(10) = 32 ft.

1.11 length and perimeter w

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