Beginning Algebra, 11th Edition

SECTION 2.5 Formulas and Additional Applications from Geometry^121

OBJECTIVE 2 Use a formula to solve an applied problem. When solving an

applied problem that involves a geometric figure, it is a good idea to draw a sketch.

Examples 2 and 3use the idea of perimeter. The perimeterof a plane (two-dimensional)

geometric figure is the distance around the figure. For a polygon (e.g., a rectangle,

square, or triangle), it is the sum of the lengths of its sides.

Finding the Dimensions of a Rectangular Yard

Cathleen Horne’s backyard is in the shape of a

rectangle. The length is 5 m less than twice the

width, and the perimeter is 80 m. Find the dimen-

sions of the yard.

Step 1 Read the problem. We must find the

dimensions of the yard.

Step 2 Assign a variable.Let the width

of the lot, in meters. Since the length is

5 meters less than twice the width, the

length is. See FIGURE 9.

Step 3 Write an equation.Use the formula for

the perimeter of a rectangle.

Perimeter of a rectangle Perimeter

80 2 W Substitute for length L.

Step 4 Solve. Distributive property

Combine like terms. Add 10. Combine like terms.

Divide by 6.

Step 5 State the answer.The width is 15 m and the length is

Step 6 Check.If the width is 15 m and the length is 25 m, the perimeter is

m, as required. NOW TRY

Finding the Dimensions of a Triangle

The longest side of a triangle is 3 ft longer than the shortest side. The medium side

is 1 ft longer than the shortest side. If the perimeter of the triangle is 16 ft, what are

the lengths of the three sides?

Step 1 Readthe problem. We must find the lengths of the sides of a triangle.

Step 2 Assign a variable.

Let the length of the shortest side, in feet,

the length of the medium side, in feet, and,

the length of the longest side in feet.

See FIGURE 10.

s + 3 =

s + 1 =

s=

EXAMPLE 3

21252 + 21152 = 50 + 30 = 80

21152 - 5 = 25 m.

15 =W

90

6

=

6 W

6

90 = 6 W

80 + 10 = 6 W- 10 + 10

80 = 6 W- 10

80 = 4 W- 10 + 2 W

= 212 W- 52 + 2 W- 5

= 2 #Length + 2 #Width

P= 2 L+ 2 W

L= 2 W- 5

W=

NOW TRY EXAMPLE 2

EXERCISE 2
Kurt’s garden is in the shape
of a rectangle. The length is
10 ft less than twice the
width, and the perimeter is
160 ft. Find the dimensions
of the garden.

W

2 W – 5

FIGURE 9

s s + 1

s + 3 FIGURE 10

NOW TRY ANSWER

width: 30 ft; length: 50 ft