1.12 area w
73
Area http://www.lahc.edu/math/frankma.htm
Transcript of 1.12 area w
AreaIf we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.
AreaIf 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.
AreaIf 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.
AreaIf 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
1 m
1 m
1 mi
1 mi
AreaIf 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
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 m
1 m
1 mi
1 mi
AreaIf 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
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 m
1 m
1 mi
1 mi
AreaIf 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
1 in2
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 m
1 m
1 mi
1 mi
1 square-inch
AreaIf 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
1 in2
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 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile
2 mi
3 miAreaA 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).
2 mi
3 miAreaA 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).
= 6 mi2 2 x 3
2 mi
3 miAreaA 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
then its area A = h x w (unit2).
* For our discussion, the “width” is the horizontal length.
2 mi
3 miAreaA 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).
* For our discussion, the “width” is the horizontal length.
2 mi
3 miAreaA 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).
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.
2 mi
3 miAreaA 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).
A square is a rectangle with four equal sides s as shown.
s
sThe perimeter of a square is s + s + s + s = 4s.The area of a square is s*s = s2.
A square
* For our discussion, the “width” is the horizontal length.
2 mi
3 miAreaA 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).
A square is a rectangle with four equal sides s as shown.
s
sThe perimeter of a square is s + s + s + s = 4s.The area of a square is s*s = s2.So the perimeter of a 10 m x 10 m square is 40 m.and its area is 102 = 100 m2.
A square
* For our discussion, the “width” is the horizontal length.
2 mi
3 miAreaA 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).
A square is a rectangle with four equal sides s as shown.
s
sThe perimeter of a square is s + s + s + s = 4s.The area of a square is s*s = s2.So the perimeter of a 10 m x 10 m square is 40 m.and its area is 102 = 100 m2.An area with four equal sides in general is called a rhombus (diamond shapes).
s
s
A square
A rhombus* For our discussion, the “width” is the horizontal length.
2 mi
3 miAreaA 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)
* For our discussion, the “width” is the horizontal length.
then its area A = h x w (unit2). A square is a rectangle with four equal sides s as shown.
s
sThe perimeter of a square is s + s + s + s = 4s.The area of a square is s*s = s2.So the perimeter of a 10 m x 10 m square is 40 m.and its area is 102 = 100 m2.An area with four equal sides in general is called a rhombus (diamond shapes).
s
s
A square
A rhombus
The perimeter of a rhombus is 4s, but its area depends on its shape.
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
R
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
R
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
RThere are two basic approaches.
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches. R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
R
12
4
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches. R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
Hence the area of R is
R
144 – 32= 112 m2
12
4
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches.
Il. We may dissect R into two rectangles I and II as shown.
R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
1212
4 4
Hence the area of R is
R
144 – 32= 112 m2
12
4
8
I II
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches.
Il. We may dissect R into two rectangles I and II as shown.
R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
1212
4 4
Hence the area of R is
R
144 – 32= 112 m2
12
4
8
I II
Area of I is 12 x 8 = 96,area of II is 4 x 4 = 16.
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches.
Il. We may dissect R into two rectangles I and II as shown.
R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
1212
4 4
Hence the area of R is
R
144 – 32= 112 m2
12
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.
Example A. Find the area of the following shape R. Assume the unit is meter.
Area
1212
44
There are two basic approaches.
Il. We may dissect R into two rectangles I and II as shown.
R
12
8
I. We may view R as a 12 x 12 = 144 m2 square with a 4 x 8 = 32 m2
corner removed.
4
1212
4 4
Hence the area of R is
R
144 – 32= 112 m2
12
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.
1212
4 4
8
iii
iv
(We may also cut R into iii and iv as shown here.)
b. Find the area of the following shape R.Area
2 ftLet’s cut R into three rectangles as shown.
Areab. Find the area of the following shape R.
Let’s cut R into three rectangles as shown.
2 ftI
II
III
Areab. 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,
2 ftI
II
III
Areab. 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,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
Areab. 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,
2 ftI
II
IIIarea of II is 2 x 6 = 12,and area of III is 2 x 5 = 10.
Areab. 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,
2 ftI
II
IIIarea 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.
Areab. 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,
2 ftI
II
IIIarea 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?
6 ft
4 ft
25 ft
20 ft
Areab. 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,
2 ftI
II
IIIarea 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?
6 ft
4 ft
25 ft The area of the larger strip is 25 x 6 = 150 ft2
20 ft
Areab. 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,
2 ftI
II
IIIarea 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?
6 ft
4 ft
25 ft The area of the larger strip is 25 x 6 = 150 ft2 and the area of the smaller strip is 20 x 4 = 80 ft2.
20 ft
Areab. 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,
2 ftI
II
IIIarea 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?
6 ft
4 ft
25 ft The area of the larger strip is 25 x 6 = 150 ft2 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.
20 ft
Areab. 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,
2 ftI
II
IIIarea 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 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.
6 ft
4 ft
25 ft
20 ft
AreaA parallelogram is a shape enclosed by two sets of parallel lines.
hb
AreaA parallelogram is a shape enclosed by two sets of parallel lines.
hb
By cutting and pasting, we may arrange a parallelogram into a rectangle.
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
AreaA 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 = b x h whereb = base and h = height.
hb
hb
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
12 ft8 ft
Hence the area of the parallelogram is A = b x h whereb = base and h = height.
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
12 ft8 ft 12 ft8 ft
Hence the area of the parallelogram is A = b x h whereb = base and h = height.
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
12 ft8 ft
8 ft12 ft
12 ft8 ft
Hence the area of the parallelogram is A = b x h whereb = base and h = height.
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
12 ft8 ft
8 ft8 ft12 ft12 ft
12 ft8 ft
Hence the area of the parallelogram is A = b x h whereb = base and h = height.
AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.
hb
hb
For example, the area of all the parallelograms shown here is 8 x 12 = 96 ft2, so they are the same size.
12 ft8 ft
8 ft8 ft12 ft12 ft
12 ft8 ft
Hence the area of the parallelogram is A = b x h whereb = base and h = height.
AreaA triangle is half of a parallelogram.
AreaA triangle is half of a parallelogram.Given a triangle, we may copy it and paste to itself to make a parallelogram,
h
b
AreaA 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
AreaA 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
AreaA 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 b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
AreaA 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
12 ft
8 ft
Therefore the area of a triangle is b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
AreaA 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
12 ft
8 ft
12 ft
8 ft
Therefore the area of a triangle is b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
AreaA 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
12 ft
8 ft
8 ft
12 ft
12 ft
8 ft
Therefore the area of a triangle is b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
AreaA 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
12 ft
8 ft
8 ft8 ft
12 ft12 ft
12 ft
8 ft
Therefore the area of a triangle is b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
AreaA 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
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
12 ft
12 ft
8 ft
Therefore the area of a triangle is b x h
2A = (b x h) ÷ 2 or A =
where b = base and h = height.
8 ft
12 ft
AreaA trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
Example B. 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.
125
8
Area
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.
125
8A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
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.
125
8
8
The parallelogram has base = 8 and height = 5,
A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
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.
125
8
8
The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2.
A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
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.
125
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.
A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
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.
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
5
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.
A trapezoid is a 4-sided figure with one set of opposite sides parallel.
Area
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.
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
5
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 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.
A trapezoid is a 4-sided figure with one set of opposite sides parallel.
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
aT
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
T
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
a
b
a + b
The base of the parallelogram is (a + b)
T
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
a
b
a + b
The base of the parallelogram is (a + b) so its area is (a + b)h.
T
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
a
b
a + b
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
T = (a + b)h ÷ 2
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
a
b
a + b
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
T = (a + b)h ÷ 2
For example, applying this formula in the last example, we have the same answer:
125
8
AreaSuppose the lengths of the opposite parallel sides a and b and the height h of a trapezoid T are known.
bh
a
By pasting another copy of the Tto itself, we obtain a parallelogram.
bh
a
a
b
a + b
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
T = (a + b)h ÷ 2
For example, applying this formula in the last example, we have the same answer:(12 + 8) 5÷2 = 100÷2 = 50.
125
8
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