Activity: Solve the followingproblems involving quadraticequations. Show yourcomplete solution.1.A rectangular garden has anarea of 84m² and a perimeterof 38m. Find its length andwidth.2. The area of a concretepathway is 350 m² and itsperimeter pathway is 90 m.What is the length of thepathway?
Answer (1)
Answer:Here are the solutions to the problems: Problem 1: Rectangular Garden Let: - l = length of the garden- w = width of the garden Given: - Area = l * w = 84 m² - Perimeter = 2l + 2w = 38 m Solution: 1. Solve for one variable in terms of the other:From the perimeter equation, we can solve for l : 2l = 38 - 2w l = 19 - w 2. Substitute into the area equation:Substitute the value of l from step 1 into the area equation: (19 - w) * w = 84 3. Simplify and solve the quadratic equation: 19w - w² = 84 w² - 19w + 84 = 0 Factor the quadratic equation: (w - 12)(w - 7) = 0 Therefore, w = 12 or w = 7 4. Find the length:If w = 12 , then l = 19 - 12 = 7 If w = 7 , then l = 19 - 7 = 12 Answer: The length of the garden is 12 meters and the width is 7 meters (or vice versa). Problem 2: Concrete Pathway Let: - l = length of the pathway- w = width of the pathway Given: - Area = l * w = 350 m² - Perimeter = 2l + 2w = 90 m Solution: 1. Solve for one variable in terms of the other:From the perimeter equation, we can solve for l : 2l = 90 - 2w l = 45 - w 2. Substitute into the area equation:Substitute the value of l from step 1 into the area equation: (45 - w) * w = 350 3. Simplify and solve the quadratic equation: 45w - w² = 350 w² - 45w + 350 = 0 Factor the quadratic equation: (w - 10)(w - 35) = 0 Therefore, w = 10 or w = 35 4. Find the length:If w = 10 , then l = 45 - 10 = 35 If w = 35 , then l = 45 - 35 = 10 Answer: The length of the pathway is 35 meters and the width is 10 meters (or vice versa).
