The document discusses basic geometric shapes and formulas for calculating their perimeters. It defines a loop or polygon as a shape formed by connecting line segments end to end. Triangles have three sides and their perimeter is calculated as the sum of the three side lengths. Specific types of triangles like equilateral triangles are discussed. Rectangles are four-sided polygons with right angles, and squares are rectangles with four equal sides. Formulas are provided for calculating the perimeters of squares and rectangles based on their side lengths. Some example problems demonstrate applying these concepts and formulas to calculate perimeters of fenced or roped areas composed of multiple shapes.

1. Basic Geometrical Shapes and Formulas

2. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
Basic Geometrical Shapes and Formulas

3. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
Basic Geometrical Shapes and Formulas

4. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
Basic Geometrical Shapes and Formulas

5. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
Basic Geometrical Shapes and Formulas

6. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.
Basic Geometrical Shapes and Formulas

7. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.
Basic Geometrical Shapes and Formulas

8. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.
All the areas above are enclosed by the same rope,
so they have equal perimeters.
Basic Geometrical Shapes and Formulas

9. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.
All the areas above are enclosed by the same rope,
so they have equal perimeters.
Following shapes are polygons:
A plane-shape is a polygon if it is formed by straight lines.
Basic Geometrical Shapes and Formulas

10. If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.
All the areas above are enclosed by the same rope,
so they have equal perimeters.
Following shapes are polygons: These are not polygons:
A plane-shape is a polygon if it is formed by straight lines.
Basic Geometrical Shapes and Formulas

11. Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas

12. Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c

13. Triangles with three equal sides are
equilateral triangles.
Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c

14. Triangles with three equal sides are
equilateral triangles.
Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c
The perimeter of an equilateral triangle is P = 3s.

15. Triangles with three equal sides are
equilateral triangles.
Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c
The perimeter of an equilateral triangle is P = 3s.
Rectangles are 4-sided
polygons where the sides
are joint at a right angle as
shown.

16. Triangles with three equal sides are
equilateral triangles.
Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c
The perimeter of an equilateral triangle is P = 3s.
Rectangles are 4-sided
polygons where the sides
are joint at a right angle as
shown. s
s
ss
A square
Rectangle with four equal sides are squares.

17. Triangles with three equal sides are
equilateral triangles.
Three sided polygons
are triangles.
Basic Geometrical Shapes and Formulas
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c
The perimeter of an equilateral triangle is P = 3s.
Rectangles are 4-sided
polygons where the sides
are joint at a right angle as
shown. s
s
ss
A square
Rectangle with four equal sides are squares.
The perimeter of a squares is
P = s + s + s + s = 4s

18. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft

19. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.

20. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.
There are 9 sections where each is 20 ft hence we need
9 x 20 = 180 ft of fence.

21. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.
There are 9 sections where each is 20 ft hence we need
9 x 20 = 180 ft of fence.
If we know two adjacent sides of a rectangle, then we know
all four sides because their opposites sides are identical.

22. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.
There are 9 sections where each is 20 ft hence we need
9 x 20 = 180 ft of fence.
If we know two adjacent sides of a rectangle, then we know
all four sides because their opposites sides are identical.
We will use the word “height” for
the vertical side and “width” for the
horizontal side.
width (w)
height (h)

23. Basic Geometrical Shapes and Formulas
Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.
There are 9 sections where each is 20 ft hence we need
9 x 20 = 180 ft of fence.
If we know two adjacent sides of a rectangle, then we know
all four sides because their opposites sides are identical.
We will use the word “height” for
the vertical side and “width” for the
horizontal side. The perimeter of a
rectangle is h + h + w + w or that
width (w)
height (h)
P = 2h + 2w

24. Example B. a. 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?
Basic Geometrical Shapes and Formulas
50 m
70 m

25. Example B. a. 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?
Basic Geometrical Shapes and Formulas
50 m
70 m
We have three heights where each
requires 50 meters of rope,

26. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,

27. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,

28. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,
b. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft
The perimeter of the step-shape is the
same as the perimeter of the rectangle
that boxes it in as shown.

29. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,
b. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft

30. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,
b. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft
The perimeter of the step-shape is the
same as the perimeter of the rectangle
that boxes it in as shown.
2 ft

31. Example B. a. 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?
Basic Geometrical Shapes and Formulas
and three widths where each
requires 70 meters of rope.
Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
We have three heights where each
requires 50 meters of rope,
b. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft
The perimeter of the step-shape is the
same as the perimeter of the rectangle
that boxes it in as shown.
2 ft
The height of the rectangle is 6 ft and the width is 10 ft, so the
perimeter P = 2(6) +2(10) = 32 ft.

32. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.

33. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.

34. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.

35. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.

36. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.

37. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.
1 in
1 in
1 in2
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 square-inch

38. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.
1 in
1 in
1 in2
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 m2
1 square-inch 1 square-meter

39. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.
1 in
1 in
1 in2
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile

40. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
the rope form a loop
that encloses an area.
The word “area”
also denotes the amount of surface enclosed.
1 in
1 in
1 in2
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile
We find the area of rectangles by cutting them into squares.

41. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
w
= 6 mi2
2 x 3

42. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).

43. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.

44. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.

45. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
R
12 12

46. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
There are two basic approaches.
R
12 12

47. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
There are two basic approaches.
R
I. We may view R as a 12 x 12 square
with a 4 x 8 corner removed.
12
8
12
4
4
R
12 12

48. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
There are two basic approaches.
R
I. We may view R as a 12 x 12 square
with a 4 x 8 corner removed.
12
8
12
4
4
R
12 12
Hence the area of R is
12 x 12 – 4 x 8

49. 2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).
The area of a square is s*s = s2.
By cutting and pasting, we may find areas of other shapes.
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
There are two basic approaches.
R
I. We may view R as a 12 x 12 square
with a 4 x 8 corner removed.
12
8
12
4
4
R
12 12
Hence the area of R is
12 x 12 – 4 x 8 = 144 – 32 = 112 m2.

50. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
I II

51. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,

52. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.

53. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.

54. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
2 ft

55. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
2 ft

56. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
I
II
III 2 ft

57. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
I
II
III 2 ft

58. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
area of II is 2 x 6 = 12,
I
II
III 2 ft

59. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
area of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
I
II
III 2 ft

60. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
area of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
I
II
III 2 ft

61. Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
area of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
I
II
III 2 ft
By cutting and pasting we obtain the following area formulas.

62. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
h
b

63. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
h
b

64. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
h
b

65. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
h
b
h
b

66. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
Hence the area of the parallelogram is A = h x b where
h = height and b = base.
h
b
h
b

67. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
Hence the area of the parallelogram is A = h x b where
h = height and b = base.
h
b
h
b
For example, the area of all
the parallelograms shown
here is 8 x 12 = 96 ft2,
so they are the same size.
12 ft
8 ft

68. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
Hence the area of the parallelogram is A = h x b where
h = height and b = base.
h
b
h
b
For example, the area of all
the parallelograms shown
here is 8 x 12 = 96 ft2,
so they are the same size.
12 ft
8 ft 12 ft8 ft

69. Area
A parallelogram is a shape enclosed by two sets of parallel lines.
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
Hence the area of the parallelogram is A = h x b where
h = height and b = base.
h
b
h
b
For example, the area of all
the parallelograms shown
here is 8 x 12 = 96 ft2,
so they are the same size.
12 ft
8 ft
8 ft8 ft
12 ft12 ft
12 ft8 ft

70. Area
A triangle is half of a parallelogram.

71. Area
A triangle is half of a parallelogram.

72. Area
A triangle is half of a parallelogram.
Given a triangle, we may copy it
and paste to itself to make a
parallelogram,
h
b
h
b

73. Area
A triangle is half of a parallelogram.
Given a triangle, we may copy it
and paste to itself to make a
parallelogram, and the area of the
triangle is half of the area of the
parallelogram formed.
h
b
h
b

74. Area
A triangle is half of a parallelogram.
Given a triangle, we may copy it
and paste to itself to make a
parallelogram, and the area of the
triangle is half of the area of the
parallelogram formed.
h
b
h
b
Therefore the area of a triangle is
h x b
2
A = (h x b) ÷ 2 or A =
where h = height and b = base.

75. Area
A triangle is half of a parallelogram.
Given a triangle, we may copy it
and paste to itself to make a
parallelogram, and the area of the
triangle is half of the area of the
parallelogram formed.
h
b
h
b
Therefore the area of a triangle is
h x b
2
A = (h x b) ÷ 2 or A =
where h = height and b = base.
For example, the area of all
the triangles shown here is
(8 x 12) ÷ 2 = 48 ft2,
i.e. they are the same size.12 ft
8 ft 8 ft

76. Area
A triangle is half of a parallelogram.
Given a triangle, we may copy it
and paste to itself to make a
parallelogram, and the area of the
triangle is half of the area of the
parallelogram formed.
h
b
h
b
Therefore the area of a triangle is
h x b
2
A = (h x b) ÷ 2 or A =
where h = height and b = base.
For example, the area of all
the triangles shown here is
(8 x 12) ÷ 2 = 48 ft2,
i.e. they are the same size.12 ft
8 ft
8 ft8 ft
12 ft12 ft
12 ft
8 ft

77. Area
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.

78. Area
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
12
5
8
R

79. Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
8
84
5
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
R

80. Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
8
84
The parallelogram has base = 8 and height = 5,
hence its area is 8 x 5 = 40 m2.
5
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
R

81. Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
8
84
The parallelogram has base = 8 and height = 5,
hence its area is 8 x 5 = 40 m2.
The triangle has base = 4 and height = 5,
hence its area is (4 x 5) ÷ 2 = 10 m2.
5
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
R

82. Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
8
84
The parallelogram has base = 8 and height = 5,
hence its area is 8 x 5 = 40 m2.
The triangle has base = 4 and height = 5,
hence its area is (4 x 5) ÷ 2 = 10 m2.
5
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
R

83. Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
8
84
The parallelogram has base = 8 and height = 5,
hence its area is 8 x 5 = 40 m2.
The triangle has base = 4 and height = 5,
hence its area is (4 x 5) ÷ 2 = 10 m2.
We may find the area of any trapezoid by cutting it into one
parallelogram and one triangle.
5
Example D. Find the area of the
following trapezoid R.
Assume the unit is meter.
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.
R

84. Circumference and Area of Circles
A circle has a center x,
x

85. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x

86. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.

87. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.

88. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.
Hence d = 2r.

89. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.
Hence d = 2r.
The perimeter C of a circle is called
the circumference and
C = πd or C = 2πr where π ≈ 3.14…

90. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.
Hence d = 2r.
The perimeter C of a circle is called
the circumference and
C = πd or C = 2πr where π ≈ 3.14…
We may use 3 as an under–estimation for π.

91. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.
Hence d = 2r.
The perimeter C of a circle is called
the circumference and
C = πd or C = 2πr where π ≈ 3.14…
We may use 3 as an under–estimation for π.
Example D. Is 25 feet of rope enough
to mark off a circle of radius r = 9 ft on the ground?

92. Circumference and Area of Circles
A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x
rr x
d (diameter)
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.
Hence d = 2r.
The perimeter C of a circle is called
the circumference and
C = πd or C = 2πr where π ≈ 3.14…
We may use 3 as an under–estimation for π.
Example D. Is 25 feet of rope enough
to mark off a circle of radius r = 9 ft on the ground?
No, 25 ft is not enough since the circumference C is at least
3 x 9 = 27 ft.

93. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
A

94. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
A

95. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
Estimating using 3 in stead of 3.14 …
we have the area A to be at least
3 x 52
A

96. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
Estimating using 3 in stead of 3.14 …
we have the area A to be at least
3 x 52 = 3 x 25 = 75 m2.
A

97. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
Estimating using 3 in stead of 3.14 …
we have the area A to be at least
3 x 52 = 3 x 25 = 75 m2.
b. Approximate the area of the circle with a 5–meter radius
using π = 3.14 as the estimated value of π.
A

98. Circumference and Area of Circles
r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
The better approximate answer using π = 3.14
is 3.14 x 52 = 3.14 x 75 = 78.5 m2
Estimating using 3 in stead of 3.14 …
we have the area A to be at least
3 x 52 = 3 x 25 = 75 m2.
b. Approximate the area of the circle with a 5–meter radius
using π = 3.14 as the estimated value of π.
A

99. Volume
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.

100. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube

101. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube

102. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in

103. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m

104. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m
1 mi 1 mi
1 mi31 mi
1 cubic mile

105. Volume
s
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m
1 mi 1 mi
1 mi31 mi
1 cubic mile
We can cut larger cubes into smaller cubes to calculate their
volume.

106. Volume
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

107. Volume
w = width
A rectangular box is specified by three sides:
the length, the width, and the height.
We say that the dimension of the box
is “l by w by h”. l = length
h = height
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

108. Volume
w = width
A rectangular box is specified by three sides:
the length, the width, and the height.
We say that the dimension of the box
is “l by w by h”. l = length
h = height
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
Here is a “4 by 3 by 2” box.
4 3
2
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

109. Volume
w = width
A rectangular box is specified by three sides:
the length, the width, and the height.
We say that the dimension of the box
is “l by w by h”. l = length
h = height
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
Here is a “4 by 3 by 2” box.
Assuming the unit is inch, then the box
may be cut into 2 x 3 x 4 = 24
1–inch cubes so its volume is 24 in3. 4 3
2
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

110. Volume
w = width
A rectangular box is specified by three sides:
the length, the width, and the height.
We say that the dimension of the box
is “l by w by h”. l = length
h = height
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
Here is a “4 by 3 by 2” box.
Assuming the unit is inch, then the box
may be cut into 2 x 3 x 4 = 24
1–inch cubes so its volume is 24 in3.
We define the volume V of a box whose sides are l, w, and h
to be V = l x w x h unit3.
In particular the volume of a cube whose sides equal to s is
V = s x s x s = s3 unit3.
4 3
2
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

111. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)

112. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
One cubic foot is
1 ft x 1 ft x 1ft

113. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.

114. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
Example b. How many cubic feet are there in the following
solid?
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.

115. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
Example b. How many cubic feet are there in the following
solid?
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.
We may view the solid is consisted of
two solids I and II as shown.
II
I
Method 1.

116. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
Example b. How many cubic feet are there in the following
solid?
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.
We may view the solid is consisted of
two solids I and II as shown.
II
I
The volume of I is 3 x 3 x 3 = 27,
the volume of II is 10 x 3 x 6 = 180.
Method 1.

117. Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)
Example b. How many cubic feet are there in the following
solid?
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.
We may view the solid is consisted of
two solids I and II as shown.
II
I
The volume of I is 3 x 3 x 3 = 27,
the volume of II is 10 x 3 x 6 = 180.
Hence the volume of the entire solid is
180 + 27 = 207 ft3.
Method 1.

118. Volume
The solid may be viewed as a box
with volume 3 x 10 x 9 = 270
with a top portion removed.
Method 2.
9 ft
10 ft
3 ft

119. Volume
The solid may be viewed as a box
with volume 3 x 10 x 9 = 270
with a top portion removed.
The dimension of the removed portion
is also a box with volume 3 x 3 x 7 = 63.
Method 2.
9 ft
10 ft
3 ft
3 ft
3 ft
7 ft

120. Volume
The solid may be viewed as a box
with volume 3 x 10 x 9 = 270
with a top portion removed.
The dimension of the removed portion
is also a box with volume 3 x 3 x 7 = 63.
Hence the volume of the given solid is
270 – 63 = 207 ft3.
Method 2.
9 ft
10 ft
3 ft
3 ft
3 ft
7 ft