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Example 3.45 from Elementary Algebra, 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem: example 3.45
The width of a rectangle is two feet less than the length. The perimeter is 52 feet. Find the length and width.
Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.
Solution
Step 1. Read the problem. Step 2. Identify what you are looking for. the length and width of a rectangle Step 3. Name. Choose a variable to represent it.
Since the width is defined in terms of the length, we let L = length. The width is two feet less than the length, so we let L − 2 = width.
ftStep 4. Translate. Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. Substitute in the given information. Step 5. Solve the equation. Combine like terms. Add 4 to each side. Divide by 4.
The length is 14 feet.Now we need to find the width. The width is .
The width is 12 feet.Step 6. Check.
Since , this works!Step 7. Answer the question. The length is 14 feet and the width is 12 feet.
Your turn
The book's Try It right after this example: same idea, new numbers. Only the answer is given.
The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.
How did it go?