# Geometric Word Problems

Geometric Word Problems

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?

Solution

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

1. 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?

Solution

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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

1. 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.

Solution

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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)

1. 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.

Solution

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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.

1. 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.

Solution

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.

1. 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.

Solution

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

1. 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

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Let x = number of degrees in first angle

x + 20 = number of degrees in second angle

x + 10 = number of degrees in third angle

Equation

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

4x + 60 = 180

4x = 180

x = 30

x + 20 = 50

2(x + 20) = 100

So, the three sides are 30, 50, and 100 degrees.

1. 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

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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.

120 + 120 + 40 = 280 feet of fence

1. 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

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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

So, the three sides are 21, 42, and 43 degrees.

1. 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 room’s dimensions?

Solution

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Let x = width in feet.

2x + 8 = length in feet.

Equation

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

124 = 6x + 4x + 16

124 = 6x + 16

108 = 6x

x = 18

2x + 8 = 44

So, The width is 18 and the length is 44

1. 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

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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

So, the three sides are 18, 13, and 11 degrees.

1. 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

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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

1. 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

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Let x = depth of box in inches.

Equation

(7)(12)(x) = 252

84x = 252

x = 3

So, the box is 3 inches deep.

1. 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

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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

# Equation

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

1. The length of a rectangle is 4 m more than the width. The area of the rectangle is 45 m2. 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 m2.

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

1. The length of a rectangle is three times the width. The area is 108 cm2. Find the dimensions of the rectangle.

Solution

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

3x = length of rectangle in feet

The area is 108 cm2

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

1. The length of a photograph is 1 cm less than twice the width. The area is 28 cm2. 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 cm2.

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

1. 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 m2. Find the length of a side of the original field.

Solution

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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 m2.

(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

1. The length of a rectangular mural is 1m greater than the width. The area is 20 m2. 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 m2.

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

1. The length of a rectangle is 6 cm more than the width. The area is 11 cm2. 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 m2.

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

1. The length of a rectangular garden is 4 m greater than the width. The area is 77 m2. 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 m2.

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

1. The length of a rectangular park is 2 km less than twice the width. The area is 112 km2. 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 km2.

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