Problem

Solve the problem. See Examples.EXAMPLE Finding the Dimensions of a Rectangular YardCathle...

Solve the problem. See Examples.

EXAMPLE 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 W = the width of the lot, in meters. Since the length is 5 meters less than twice the width, the length is L = 2W − 5. See FIGURE 9.
FIGURE 9
Step 3 Write an equation. Use the formula for the perimeter of a rectangle.
Step 4 Solve.
Step 5 State the answer. The width is 15 m and the length is 2(15) − 5 = 25 m.
Step 6 Check. If the width is 15 m and the length is 25 m, the perimeter is 2(25) + 2(15) = 50 + 30 = 80 m, as required.
EXAMPLE 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 Read the problem. We must find the lengths of the sides of a triangle.
Step 2 Assign a variable.
See FIGURE 10.
FIGURE 10
Step 3 Write an equation. Use the formula for the perimeter of a triangle.
Step 4
Step 5 State the answer. The shortest side, s, has length 4 ft. Then
Step 6 Check. The medium side, 5 ft, is 1 ft longer than the shortest side, and the longest side, 7 ft, is 3 ft longer than the shortest side. Futhermore, the perimeter is 4 + 5 + 7 = 16 ft, as required.