9-3 Area of Parallelograms Course 2 Warm Up Problem of the Day Lesson Presentation.
19
9-3 Area of Parallelograms Course 2 Warm Up Problem of the Day Lesson Presentation
-
Upload
christian-allison -
Category
Documents
-
view
220 -
download
0
Transcript of 9-3 Area of Parallelograms Course 2 Warm Up Problem of the Day Lesson Presentation.
9-3 Area of Parallelograms
Course 2
Warm Up
Problem of the Day
Lesson Presentation
Warm UpFind each product.
1. 8 12
2. 3
3. 9.4 6.3
4. 3.5 7
96
59.22
Course 2
9-3 Area of Parallelograms
12
5 13
18 23
24.5
Problem of the Day
How many 3 ft by 2 ft rectangles can you cut from one 8 ft by 4 ft rectangle? How much will be left over?
5 pieces; 2 ft2 left over
Course 2
9-3 Area of Parallelograms
Learn to find the area of rectangles and other parallelograms.
Course 2
9-3 Area of Parallelograms
Vocabulary
area
Insert Lesson Title Here
Course 2
9-3 Area of Parallelograms
Course 2
9-3 Area of Parallelograms
The area of a figure is the number of unit squares needed to cover the figure. Area is measured in square units.
AREA OF A RECTANGLE
The area A of a rectangle is the product of its length l and its width w.
A = lw w
l
Find the area of the rectangle.
Additional Example 1: Finding the Area of a Rectangle
Course 2
9-3 Area of Parallelograms
4.5 in.
7.4 in.
A = lw
A = 7.4 · 4.5
A = 33.3
The area of the rectangle is 33.3 in2.
Use the formula.
Substitute for l and w.
Multiply.
Find the area of the rectangle.
Course 2
9-3 Area of Parallelograms
6.3 in.
8.2 in.
A = lw
A = 8.2 · 6.3
A = 51.66
The area of the rectangle is 51.66 in2.
Use the formula.
Substitute for l and w.
Multiply.
Check It Out: Example 1
The area of a playing field is 1,470 ft2 and the length is 42 ft. What is the width of the field?
Additional Example 2: Finding Length or Width of a Rectangle
Course 2
9-3 Area of Parallelograms
A = lw
1,470 = 42 · w
= · w
The width of the playing field is 35 ft.
Use the formula for the area of a rectangle.
Substitute 1,470 for A and 42 for l.
Divide both sides by 42 to isolate w.1,47042
4242
35 = w
The area of a rectangular parking lot is 3,570 ft2 and the width is 70 ft. What is the length of the parking lot?
Check It Out: Example 2
Course 2
9-3 Area of Parallelograms
A = lw
3,570 = l · 70
= · w
The length of the parking lot is 51 ft.
Use the formula for the area of a rectangle.
Substitute 3,570 for A and 70 for w.
Divide both sides by 70 to isolate w.3,57070
7070
51 = l
Course 2
9-3 Area of Parallelograms
The area A of a parallelogramis the product of its base b and its height h.
AREA OF A PARALLELOGRAM
A = bh h
b
Find the area of the parallelogram.
Additional Example 3: Finding the Area of a Parallelogram
Course 2
9-3 Area of Parallelograms
8 m
16 m
A = bh
A = 16· 8
A = 128
The area of the parallelogram is128 m2.
Find the area of the parallelogram.
Course 2
9-3 Area of Parallelograms
6 cm
12 cm
A = bh
A = 12· 6
A = 72
The area of the parallelogram is72 cm2.
Check It Out: Example 3
A carpenter is covering a 150 ft2 floor with square tiles that are each 2 ft in length. What is the least number of tiles the carpenter will need to cover the floor?
Additional Example 4: Measurement Application
Course 2
9-3 Area of Parallelograms
First find the area of each tile.
A = lw Use the formula for the area of a square.
A = 2 · 2 Substitute 2 for l and 2 for w.
A = 4 Multiply.
The area of each square tile is 4 ft2.
Additional Example 4 Continued
Course 2
9-3 Area of Parallelograms
To find the number of tiles needed, divide the area of the floor by the area of one tile.
150 ft2
4 ft2= 37.5
Since covering the floor requires more than 37 tiles, the carpenter would need at least 38 tiles.
Check It Out: Example 4
Insert Lesson Title Here
Course 2
9-3 Area of parallelograms
First find the area of each tile.
A = lw Use the formula for the area of a square.
A = 1.5 · 1.5 Substitute 1.5 for l and 1.5 for w.
A = 2.25 Multiply.
The area of each square tile is 2.25 in2.
Amanda is covering a 200 in2 mosaic frame with square tiles that are each 1.5 inches in length. What is the least number of tiles Amanda will need to cover the frame?
Check It Out: Example 4 Continued
Insert Lesson Title Here
Course 2
9-3 Area of Parallelograms
To find the number of tiles needed, divide the area of the frame by the area of one tile.
200 ft2
2.25 ft2≈ 88.9
Since covering the frame requires more than 88 tiles, Amanda would need at least 89 tiles.
Lesson Quiz: Part I
Find the area of each figure.
1. 2.
3.
24.5 ft2
Insert Lesson Title Here
84 ft2
Course 2
9-3 Area of Parallelograms
4.
1318
in2105 8 or
28 12
ft2
572
or
Lesson Quiz: Part II
5. Suzanne is planning to use 1 ft by 0.5 ft tiles to finish her bathroom floor. If her floor is 7 ft by 10 ft, how many tiles will she need?
140 tiles
Insert Lesson Title Here
Course 2
9-3 Area of Parallelogram
