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

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

3. Area
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 an area, i.e. a surface in the plane.

4. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.

5. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
1m
1m
1 mi
1 mi

6. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
If each side of a square is 1 unit, then we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
1 in
1 in
1m
1m
1 mi
1 mi

7. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
If each side of a square is 1 unit, then 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 in
1m
1 mi
1 in
1m
1 mi

8. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
If each side of a square is 1 unit, then 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 in
1m
1 mi
1 in
1 in2
1 square-inch
1m
1 mi

9. Area
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 an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
If each side of a square is 1 unit, then 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 in
1m
1 mi
1 in
1 in2
1 square-inch
1m
1 m2
1 square-meter
1 mi
1 mi2
1 square-mile

10. Area
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).
3 mi
2 mi

11. Area
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).
3 mi
2 mi
2x3
= 6 mi2

12. Area
A 2 mi x 3 mi rectangle may be cut into
2 mi
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
2).
then its area A = h x w (unit
* For our discussion, the “width” is the horizontal length.
3 mi
2x3
= 6 mi2
w

13. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
* For our discussion, the “width” is the horizontal length.

14. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
A square is a rectangle with four equal
sides s as shown.
s
s
A square
* For our discussion, the “width” is the horizontal length.

15. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
A square is a rectangle with four equal
sides s as shown.
s
The perimeter of a square is s + s + s + s = 4s.
The area of a square is s*s = s2.
s
A square
* For our discussion, the “width” is the horizontal length.

16. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
A square is a rectangle with four equal
sides s as shown.
s
The perimeter of a square is s + s + s + s = 4s.
The area of a square is s*s = s2.
s
So the perimeter of a 10 m x 10 m square is 40 m.
A square
2 = 100 m2.
and its area is 10
* For our discussion, the “width” is the horizontal length.

17. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
A square is a rectangle with four equal
sides s as shown.
s
The perimeter of a square is s + s + s + s = 4s.
The area of a square is s*s = s2.
s
So the perimeter of a 10 m x 10 m square is 40 m.
A square
2 = 100 m2.
and its area is 10
s
An area with four equal sides in general is called
a rhombus (diamond shapes).
s
A rhombus
* For our discussion, the “width” is the horizontal length.

18. Area
3 mi
A 2 mi x 3 mi rectangle may be cut into
2x3
2 mi
six 1 x 1 squares so it covers an area of
= 6 mi2
2 x 3 = 6 mi2 (square miles).
In general, given the rectangle with
w
height = h (units)
width* = w (units),
h A = h x w (unit2)
then its area A = h x w (unit2).
A square is a rectangle with four equal
sides s as shown.
s
The perimeter of a square is s + s + s + s = 4s.
The area of a square is s*s = s2.
s
So the perimeter of a 10 m x 10 m square is 40 m.
A square
2 = 100 m2.
and its area is 10
s
An area with four equal sides in general is called
a rhombus (diamond shapes). The perimeter of
s
a rhombus is 4s, but its area depends on its shape.
A rhombus
* For our discussion, the “width” is the horizontal length.

19. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
4
R
12
12

20. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
4
R
12
12

21. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
There are two basic approaches.
4
4
R
12
12

22. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
There are two basic approaches.
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
8
4
R
12
12
4
4
4
R
12
12

23. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
There are two basic approaches.
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
Hence the area of R is
144 – 32
= 112 m2
8
4
R
12
12
4
4
4
R
12
12

24. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
R
There are two basic approaches.
12
12
I. We may view R as a
12 x 12 = 144 m2 square
Il. We may dissect R into two
2
with a 4 x 8 = 32 m
rectangles I and II as shown.
corner removed.
Hence the area of R is
144 – 32
= 112 m2
8
8
4
R
12
4
4
I
12
12
4
II
12
4

25. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
R
There are two basic approaches.
12
12
I. We may view R as a
12 x 12 = 144 m2 square
Il. We may dissect R into two
2
with a 4 x 8 = 32 m
rectangles I and II as shown.
corner removed.
Area of I is 12 x 8 = 96,
Hence the area of R is
area of II is 4 x 4 = 16.
144 – 32
= 112 m2
8
8
4
R
12
4
4
I
12
12
4
II
12
4

26. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
R
4
There are two basic approaches.
12
12
I. We may view R as a
12 x 12 = 144 m2 square
Il. We may dissect R into two
2
with a 4 x 8 = 32 m
rectangles I and II as shown.
corner removed.
Area of I is 12 x 8 = 96,
Hence the area of R is
area of II is 4 x 4 = 16.
144 – 32
The area of R is the sum of the
= 112 m2
two or 96 + 16 = 112 m2.
8
8
4
R
12
4
4
I
12
12
4
II
12

27. Area
Example A. Find the area of the following
shape R. Assume the unit is meter.
4
4
R
There are two basic approaches.
12
12
I. We may view R as a
12 x 12 = 144 m2 square
Il. We may dissect R into two
2
with a 4 x 8 = 32 m
rectangles I and II as shown.
corner removed.
Area of I is 12 x 8 = 96,
Hence the area of R is
area of II is 4 x 4 = 16.
144 – 32
The area of R is the sum of the
= 112 m2
two or 96 + 16 = 112 m2.
8
4
R
12
4
4
I
12
12
8
8
iii
4
II
12
12
4
4
iv
(We may also cut R
into iii and iv as shown here.)
12

28. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
2 ft

29. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
I
II
III
2 ft

30. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
I
II
III
2 ft

31. Area
b. Find the area of the following shape R.
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

32. Area
b. Find the area of the following shape R.
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

33. Area
b. Find the area of the following shape R.
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

34. Area
b. Find the area of the following shape R.
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.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
I
2 ft
II
III
20 ft
4 ft
25 ft
6 ft

35. Area
b. Find the area of the following shape R.
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.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
The area of the larger strip is 25 x 6 = 150 ft2
I
2 ft
II
III
20 ft
4 ft
25 ft
6 ft

36. Area
b. Find the area of the following shape R.
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
2 ft
II
III
20 ft
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
25 ft
2
The area of the larger strip is 25 x 6 = 150 ft
and the area of the smaller strip is 20 x 4 = 80 ft2.
4 ft
6 ft

37. Area
b. Find the area of the following shape R.
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
2 ft
II
III
20 ft
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
25 ft
2
The area of the larger strip is 25 x 6 = 150 ft
and the area of the smaller strip is 20 x 4 = 80 ft2.
Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2)
rectangular overlap twice.
4 ft
6 ft

38. Area
b. Find the area of the following shape R.
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
2 ft
II
III
20 ft
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
25 ft
2
The area of the larger strip is 25 x 6 = 150 ft
and the area of the smaller strip is 20 x 4 = 80 ft2.
Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2)
rectangular overlap twice. Hence the total area covered is
230 – 24 = 206 ft2.
4 ft
6 ft

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

40. 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

41. 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

42. 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
h
b
b

43. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.

44. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.
8 ft
12 ft

45. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.
8 ft
12 ft
8 ft
12 ft

46. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.
8 ft
12 ft
8 ft
8 ft
12 ft
12 ft

47. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.
8 ft
12 ft
8 ft
8 ft
12 ft
12 ft
8 ft
12 ft

48. 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
h
b
b
Hence the area of the parallelogram is A = b x h where
b = base and h = height.
For example, the area of all
the parallelograms shown
8 ft
8 ft
12 ft
here is 8 x 12 = 96 ft2,
12 ft
so they are the same size.
8 ft
12 ft
8 ft
12 ft

49. Area
A triangle is half of a parallelogram.

50. 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

51. 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

52. 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

53. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
h
b
h
b

54. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
8 ft
12 ft
h
b
h
b

55. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
8 ft
8 ft
12 ft
12 ft
h
b
h
b

56. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
8 ft
8 ft
12 ft
12 ft
8 ft
12 ft
h
b
h
b

57. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
h
b
h
b
8 ft
8 ft
12 ft
12 ft
8 ft
8 ft
12 ft
12 ft

58. 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.
Therefore the area of a triangle is
A = (b x h) ÷ 2 or A = b x h
2
where b = base and h = height.
8 ft
8 ft
12 ft
12 ft
h
b
h
b
For example, the area of all
the triangles shown here is
(8 x 12) ÷ 2 = 48 ft2,
i.e. they are the same size.
8 ft
8 ft
12 ft
12 ft

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

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

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

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

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

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

65. Area
8
A trapezoid is a 4-sided figure with
one set of opposite sides parallel. 5
12
Example B. Find the area of the
following trapezoid R.
4
8
Assume the unit is meter.
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
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.
Therefore the area of the trapezoid is 40 + 10 = 50 m2.

66. Area
8
A trapezoid is a 4-sided figure with
one set of opposite sides parallel. 5
12
Example B. Find the area of the
following trapezoid R.
4
8
Assume the unit is meter.
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
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.
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
We may find the area of any trapezoid by slicing it one
parallelogram and one triangle.
A direct formula for the area of a trapezoid may be obtained by
pasting two copies together as shown on the next slide.

67. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
a
T
h
b

68. Area
a
T
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
h
b
a
h
b

69. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
The base of the parallelogram is
(a + b)
a
T
h
b
a
b
h
b
a
a+b

70. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
The base of the parallelogram is
(a + b) so its area is (a + b)h.
a
T
h
b
a
b
h
b
a
a+b

71. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
The base of the parallelogram is
(a + b) so its area is (a + b)h.
Therefore the area of a trapezoid T
is half of (a + b)h or that
T = (a + b)h ÷ 2
a
T
h
b
a
b
h
b
a
a+b

72. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
The base of the parallelogram is
(a + b) so its area is (a + b)h.
Therefore the area of a trapezoid T
is half of (a + b)h or that
T = (a + b)h ÷ 2
a
T
h
b
a
b
h
b
a
a+b
For example, applying this formula in the last example,
we have the same answer:
8
5
12

73. Area
Suppose the lengths of the opposite
parallel sides a and b and the
height h of a trapezoid T are known.
By pasting another copy of the T
to itself, we obtain a parallelogram.
The base of the parallelogram is
(a + b) so its area is (a + b)h.
Therefore the area of a trapezoid T
is half of (a + b)h or that
T = (a + b)h ÷ 2
a
T
h
b
a
b
h
b
a
a+b
For example, applying this formula in the last example,
we have the same answer:
8
(12 + 8) 5÷2 = 100÷2 = 50.
5
12