1. Perimeter and Area Word Problems (Using One Variable) Taught by the Bestest of all the besterest who are not bestless, Mr. Peter Richard

2. Rectangles and Algebra
To solve questions involving geometric shapes
first draw a picture (If you need to)
use formulas or common sense.
A rectangle's perimeter is found by adding all
the sides together.
A rectangle’s Area is found by multiplying length times width.
Assign a variable to one of the unknown items.
Reread the problem, and then write an equation, using the items and variables
Solve the equation.
Reread the problem and check your solution

3. Rectangles with Algebra
Jan’s art teacher told her to make a frame with the length three times the width. The perimeter is 144 inches. What are the dimensions and area of the picture?

4. Rectangles with Algebra
Jan’s art teacher told her to make a frame with the length three times the width. The perimeter is 144 inches. What are the dimensions and area of the picture?
First draw and label a picture (rectangle). w
l=3w

5. Use perimeter to find width
Second choose a formula P=2L+2W
Fill in the known values =2L+2W L=3W
Simplify =2(3w)+ 2w
3W = L W

6. Use Perimeter to find width
Solve.
3W = L W

7. Problem 2
The length of a rectangle is 5 less than four times the width. The perimeter is 100 inches. Find the area.
List the unknowns: length and width
Let W = the width of the rectangle
4W - 5 = the length of the rectangle

8. Problem 2 Recall the formula for Perimeter: P = W +W + 4W - 5 + 4W - 5
Area = 11 x (4*11 – 5) =
39 x 11 = 429

9. Try This One Partner!
The Length of a rectangle is five more than three times the width. If the perimeter is
90 feet, what is the area?

10. Problem 3
A garden is in the shape of a square with a perimeter of 42 feet. The garden is surrounded by two fences. One fence is around the perimeter of the garden, while the second fence is 3 feet from the first fence, as the figure indicates. If the material used to build the two fences costs $1.28 per foot, what was the total cost of the fences?
3 FT__

11. Problem 3
4(s) + 4(s + 6) = = 108
Cost of the fence per foot X # feet of fencing = Total Cost
($1.28)(108) = $138.24
The total cost of the fencing is $
3 ft. s
S + 6

12. Problem 3 s = length of side of inner fence. Perimeter = 4s 4s = 42
The perimeter of the inner fence + the perimeter of the outer fence = total amount of fencing needed
3 ft. s
S + 6

13. Quiz Time! Homework: Page 162 # 1, 2, 3,4, 35
Do The Worksheet. Follow Directions. You will be given credit for your diagrams, your equations, and the solutions to your equations.
Homework: Page 162 # 1, 2, 3,4, 35