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Magazine Chapter 11: Problem 39
Solve each problem. A rectangle has a length \(2 \mathrm{~m}\) less than twice its width. When \(5 \mathrm{~m}\) are added to the width, the resulting figure is a square with an area of \(144 \mathrm{~m}^{2}\). Find the dimensions of the original rectangle.
Short Answer
Expert verified
The dimensions of the rectangle are width = 7 meters and length = 12 meters.
Step by step solution
01
- Understand the problem
We need to find the dimensions of a rectangle. We know that the length is 2 meters less than twice the width, and by adding 5 meters to the width, the resulting figure is a square with an area of 144 square meters.
02
- Set up the equations
Let the width of the rectangle be denoted as \(w\). Then, the length of the rectangle can be expressed as \(2w - 2\). When 5 meters are added to the width, this creates a square with side length \(w + 5\). Therefore, we have the equation for the area of the square: \[ (w + 5)^2 = 144 \]
03
- Solve for the width
Expand and solve the equation: \[ (w + 5)^2 = 144 \] \[ w^2 + 10w + 25 = 144 \] Subtract 144 from both sides: \[ w^2 + 10w + 25 - 144 = 0 \] \[ w^2 + 10w - 119 = 0 \] Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = 10\), and \(c = -119\).
04
- Apply the quadratic formula
Calculate the discriminant: \[ b^2 - 4ac = 10^2 - 4(1)(-119) = 100 + 476 = 576 \] Find the square root of the discriminant: \[ \sqrt{576} = 24 \] Now use the quadratic formula: \[ w = \frac{-10 \pm 24}{2} \] This gives two potential solutions: \[ w = \frac{14}{2} = 7 \] and \[ w = \frac{-34}{2} = -17 \]
05
- Discard the invalid solution
Since width cannot be negative, we discard \( w = -17 \), leaving us with: \[ w = 7 \]
06
- Find the length of the rectangle
Substitute \( w = 7 \) into the expression for length: \[ l = 2w - 2 = 2(7) - 2 = 14 - 2 = 12 \]
07
- Verify the solution
To confirm, if the width is 7 meters and 5 meters are added, the side length of the square is \( 7 + 5 = 12 \). The area is \( 12^2 = 144 \), which matches the problem statement. Therefore, our solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
solving quadratics
Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). In this exercise, we used a quadratic equation to find the width of a rectangle.
Quadratic equations can often be solved using the quadratic formula: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \). This formula provides the solutions to the equation by determining the values of \( x \) that satisfy the equation. In our example, we solved \( w^2 + 10w - 119 = 0 \) where \( a = 1 \), \( b = 10 \), and \( c = -119 \).
First, calculate the discriminant \( b^2 - 4ac \). If this value is positive, there will be two real solutions. We found the discriminant to be \( 576 \), yielding two possible solutions for \( w \): 7 and -17. Since a width can't be negative, we discarded -17.
Quadratic equations can often be solved using the quadratic formula: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \). This formula provides the solutions to the equation by determining the values of \( x \) that satisfy the equation. In our example, we solved \( w^2 + 10w - 119 = 0 \) where \( a = 1 \), \( b = 10 \), and \( c = -119 \).
First, calculate the discriminant \( b^2 - 4ac \). If this value is positive, there will be two real solutions. We found the discriminant to be \( 576 \), yielding two possible solutions for \( w \): 7 and -17. Since a width can't be negative, we discarded -17.
geometry problems
Geometry problems often require understanding and manipulating shapes and their properties. In this exercise, we are dealing with a rectangle and a square. A rectangle is a four-sided figure with opposite sides equal and parallel, while a square has all sides equal and parallel.
Additionally, geometry problems frequently require setting up equations that represent the given conditions. For example, we are told that adding 5 meters to the rectangle's width makes it a square. Hence, we equated \( (w + 5)^2 \) to 144, the area of the square, and used algebraic manipulation to solve for \( w \).
Visualizing the problem can also help. Imagine the rectangle with one side being modified to create a new shape. Use diagrams where possible to get a clear picture of the relationships between different elements.
Additionally, geometry problems frequently require setting up equations that represent the given conditions. For example, we are told that adding 5 meters to the rectangle's width makes it a square. Hence, we equated \( (w + 5)^2 \) to 144, the area of the square, and used algebraic manipulation to solve for \( w \).
Visualizing the problem can also help. Imagine the rectangle with one side being modified to create a new shape. Use diagrams where possible to get a clear picture of the relationships between different elements.
area of a square
The area of a square is found by squaring the length of one of its sides. In mathematical terms, if \( s \) is the side length, then the area \( A \) is given by: \( A = s^2 \). In our exercise, the modified rectangle became a square with an area of 144 square meters, so we set up the equation \( (w + 5)^2 = 144 \).
Solving for the side length, we determined that \( w + 5 = 12 \). Hence, the side length of the square is 12 meters.
Understanding this helps bridge the rectangle and square properties. It's a perfect example of how altering dimensions in geometry can simplify problem-solving.
Solving for the side length, we determined that \( w + 5 = 12 \). Hence, the side length of the square is 12 meters.
Understanding this helps bridge the rectangle and square properties. It's a perfect example of how altering dimensions in geometry can simplify problem-solving.
dimensions of a rectangle
To find a rectangle's dimensions, you need to understand its length and width. In this problem, the width (\( w \)) satisfies \( w + 5 \) creating a square, leading us to find \( w = 7 \).
The problem states that the length (\( l \)) is 2 meters less than twice the width. Plugging in our width value, we got \( l = 2w - 2 = 12 \).
Always double-check your work. Here, the rectangle originally had dimensions 7 meters by 12 meters. Adding 5 meters to 7 meters, the width matches the square's side, and calculating the square area verifies correctness: \( 12^2 = 144 \).
Remember, accurately setting up your equations and understanding how changes in one dimension affect another are crucial.
The problem states that the length (\( l \)) is 2 meters less than twice the width. Plugging in our width value, we got \( l = 2w - 2 = 12 \).
Always double-check your work. Here, the rectangle originally had dimensions 7 meters by 12 meters. Adding 5 meters to 7 meters, the width matches the square's side, and calculating the square area verifies correctness: \( 12^2 = 144 \).
Remember, accurately setting up your equations and understanding how changes in one dimension affect another are crucial.
