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**Respect Responsibility
Readiness**

1. The length of a rectangle is equal to twice the width. The perimeter is 138 feet. What is the length (in feet) of the rectangle?

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

2x = length of rectangle in feet

The perimeter equals two lengths plus two widths.

2x + 2x + x + x = 138

6x = 138

x = 23

2x = 46

**So,** the
width is 23 and the length is 46

2.
A
rectangle has a length which is 4 feet less than three times the width. The
perimeter is 224 feet. Find the absolute value of the difference of the
dimensions?

**Draw a picture**

Let x = width of rectangle in feet

3x - 4 = length of rectangle in feet (4 feet less than three times the width)

The perimeter equals two lengths plus two widths.

2(x) + 2(3x - 4) = 224

2x + 6x - 8 = 224

8x = 232

x = 29

3x - 4 = 83

**So,**
the width is 29 and the length is 83

3.
The
first angle of a triangle is twice the second, and the third is 5 degrees larger
than the first. Find the absolute value of the difference of the two largest
angles.

**Draw a picture**

Let x = number of degrees in second angle.

2x = number of degrees in first angle.

2x + 5 = number of degrees in third angle.

The sum of three angles of a triangle equals 180 degrees..

x + 2x + (2x + 5) = 180

5x + 5 = 180

5x = 75

x = 35

2x = 70

2x + 5 = 75

**So,** the
three angles are 35, 70, and 75.

Check (35 + 70 + 75 = 180)

4.
The
length of one rectangle is two times the width. If the length is decreased by 5
feet and the width is increased by 5 feet, the area is increased by 75 square
feet. Find the absolute value of the difference of the length and width.

**Draw a picture**

Let x = width of first rectangle in feet

2x = length of first rectangle in feet.

The area of the first rectangle plus 75 equals the area of the second rectangle..

**So,** the
width is 20 and the length is 40.

5.
The
first side of a triangle is 2 inches less than twice the second side. The third
side is 10 inches longer than the second side. If the perimeter is 12 feet, find
the absolute value of the difference of the two largest sides.

**Draw a picture**

Notice that the first and third sides are longer than twice the second side. So we let x equal the second side. Also, note that the perimeter is given in feet while the sides are given in inches. The perimeter must be changed to the same units (inches), and 12 feet equals 144 inches.

Let x = second side in inches.

2x - 2 = first side in inches.

x + 10 = third side in inches.

x + (2x - 2) + (x + 10) = 144

4x + 8 = 144

4x = 136

x = 34

2x - 2 = 66

x + 10 = 44

**So,** the
three sides are 34, 66, and 44.

6. The length of a rectangular soccer field is 10 feet more than twice the width. The perimeter is 320 feet. Find the length and width.

**Draw a picture**

Let x = width in feet.

2x + 5 = length of field in feet**Equation**

The perimeter equals two lengths plus two widths.

320 = 2(x) + 2(2x + 10)

320 = 2x + 4x + 20

320 = 6x + 20

300 = 6x

x = 50

2x + 10 = 110

**So,** The
width is 50 and the length is 110

7. The second angle of a
triangle is 20° greater than the first angle. The third angle is twice the
second. Find the three angles.

Solution

**Draw a picture**

Let x = number of degrees in first angle

x + 20 = number of degrees in second angle

**Equation**

x + (x + 20) + 2(x + 20) = 180

4x + 60 = 180

4x = 180

x = 30

x + 20 = 50

2(x + 20) = 100

8. A farmer wishes to fence a
rectangular area behind his barn. The barn forms one end of the rectangle, and
the length of the rectangle is three times the width. How many linear feet of
fence must the farmer buy if the perimeter of the rectangle is 320 feet?

Solution

**Draw a picture**

Let
x = width of fenced area in feet

3x = length of fenced area in feet

**Equation**

320
= 2(3x) + 2(x)

320
= 6x + 2x

320
= 8x

x
= 40

3x
= 120

**So,**
The width is 40 and the length is 120

The barn closes one end of the area. Therefore, the farmer
needs to buy two lengths plus one width of fencing.

9. The first side of a
triangle is twice the second, and the third side is 20 feet less than three
times the second. The perimeter of the triangle is 106 feet. Find the three
sides.

Solution

**Draw a picture**

**
**

The second side is the smallest

Let x = second side in feet

2x = first side in feet

3x - 20 = third side in feet

**Equation**

x + 2x + (3x - 20) = 106

6x - 20 = 106

6x = 126

x = 21

2x = 42

3x - 20 = 43

10. The length of a room is 8
feet more than twice the width. If it takes 124 feet of molding to go around the
perimeter of the room, what are the rooms dimensions?

Solution

**Draw a picture**

Let
x = width in feet.

2x
+ 8 = length in feet.

**Equation**

124
= 2(x) + 2(2x + 8)

124
= 6x + 4x

124 = 6x + 16

108
= 6x

x
= 18

2x
+ 8 = 44

11. The perimeter of a
triangular lawn is 42 yards. The first side is 5 yards less than the second, and
third side is 2 yards less than the first. What is the length of each side of
the lawn?

Solution

**Draw a picture**

The second side is the smallest

Let x = second side of lawn in yards.

x - 5 = first side of lawn in yards.

(x - 5) - 2 = third side of lawn in yards.

**Equation**

The sum of the three sides equals the perimeter.

42 = x + (x - 5) + (x 5 - 2)

42 = x + x 5 + x - 7

42 = 3x - 12

3x = 54

x = 18

x 5 = 13

x 5 2 = 11

12. One side of a rectangular
metal plate is five times as long as the other side. If the perimeter is 72
meters, what is the length of the shorter side of the plate?

Solution

**Draw a picture**

Let x = width of plate in feet

5x = length of plate in feet

**Equation**

2(5x) + 2(x) = 72

12x = 72

x = 6

5x = 30**So,**
the length is 30 and the width is 6

13. A rectangular box of cereal contains 252 cubic inches of product. It is 12 inches tall and 7 inches wide. How deep is the box? (Volume equals height times width times depth)

Solution

**Draw a picture**

Let x = depth of box in inches.

**Equation**

(7)(12)(x) = 252

84x = 252

x = 3

14. The length of a rectangle is 8 feet more than the width. If the width is increased by 4 feet and the length is decreased by 5 feet, the area of the two rectangles remains the same. Find the dimensions of the original rectangle.

Solution

**Draw a picture**

Let x = width of original rectangle in feet.

x + 8 = length of original rectangle in feet.

Then x + 4 = width of new rectangle.

(x + 8) 5 = length of new
rectangle

The area of the first
rectangle, x(x + 8), equals the area of the second rectangle, (x + 4)(x + 3).

x(x + 8) = (x + 4)(x + 3)

FINISH

15. The length of a
rectangle is 4 m more than the width. The area of the rectangle is 45 m^{2}.
Find the length and width.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

x + 4 = length of rectangle in feet

The area is
45 m^{2}.

x(x + 4) = 45

x^2 + 4x = 45

x^2 + 4x - 45 = 0

(x - 5) (x + 9) = 0

x = 5 or -9

We must reject -9 because the length and width must be positive.

**So,**

The width is 5 and the length is 9

Check

5 x 9 = 45

16. The length of a
rectangle is three times the width. The area is 108 cm^{2}. Find the
dimensions of the rectangle.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

3x = length of rectangle in feet

The area is 108 cm^{2}

x(3x) = 108

3x^2 = 108

x^2 = 36

x = 6 or -6

We must reject -6 because the length and width must be positive.

**So,**

The width is 6 and the length is 18

Check

6(18) =108

17. The length of a
photograph is 1 cm less than twice the width. The area is 28 cm^{2}.
Find the dimensions of the photograph.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

2x + 1 = length of rectangle in feet

The area is 28 cm^{2}.

x(2x + 1) = 28

2x^2 + x = 28

2x^2 + x - 28 = 0

(2x - 7) (x + 4) = 0

x = 3.5 or -4

We must reject -4 because the length and width must be positive.

**So,**

The width is 3.5 and the length is 8

Check

3.5 x 8 = 28

18. A square field had 3 m
added to its length and 2 m added to its width. The field then had an area of 90
m^{2}. Find the length of a side of the original field.

Solution

**Draw a picture**

Let x = width of square in feet (smaller number)

x = length of square in feet

x + 2 = new width

x + 3 = new length

The field then had an area of 90 m^{2}.

(x + 2) (x + 3) = 90

x^2 + 5x + 6 = 90

x^2 + 5x - 84 = 0

(x - 7) (x + 12) = 0

x = 7 or -12

We must reject -12 because the length and width must be positive.

**So,**

The width is 9 and the length is 10

Check

9 x 10 = 90

19. The length of a
rectangular mural is 1m greater than the width. The area is 20 m^{2}.
Find the dimensions of the mural.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

x + 1 = length of rectangle in feet

The area is 20 m^{2}.

x (x + 1) = 20

x^2 + x = 20

x^2 + x - 18 = 0

(x + 5) (x - 4) = 0

x = -5 or 4

We must reject -5 because the length and width must be positive.

**So,** the
width is 4 and the length is 5

Check

4(5) = 20

20. The length of a
rectangle is 6 cm more than the width. The area is 11 cm^{2}. Find the
length and width.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

x + 1 = length of rectangle in feet

The area is 20 m^{2}.

x (x + 1) = 20

x^2 + x = 20

x^2 + x - 20 = 0

(x + 5) (x - 4) = 0

x = -5 or 4

We must reject -5 because the length and width must be positive.

**So,** the
width is 4 and the length is 5

Check

4(5) = 20

21. The length of a
rectangular garden is 4 m greater than the width. The area is 77 m^{2}.
Find the dimensions of the garden.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

x + 4 = length of rectangle in feet

The area is 77 m^{2}.

x (x + 4) = 77

x^2 + 4x = 77

x^2 + 4x - 77 = 0

(x - 7) (x +11) = 0

x = 7 or -11

We must reject -11 because the length and width must be positive.

**So,** the
width is 7 and the length is 11

Check

7(11) = 77

^{2}.^{
} Find the dimensions of the park.

Solution

**Draw a picture**

Let x = width of rectangle in feet (smaller number)

2x + 2 = length of rectangle in feet

The area is 112 km^{2}.

x (2x + 2) = 112

2x^2 + 2x = 112

2x^2 + 2x - 112 = 0

(2x +16) (x - 7) = 0

x = 7 or -8

We must reject -8 because the length and width must be positive.

**So,** the
width is 7 and the length is 16

Check

7(16) = 112