Can someone help me set up the equations thanks.
Directions: Solve each of the following applications. Give all answers to the nearest thousandth.
Problem: Geometry. The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
Let x = width of rectangle
Let x + 1 = length of rectangle
Remember that a^2 + b^2 = c^2
Therefore:
x^2 + (x + 1)^2 = 4^2
Can you take it from here to solve for x?
Don't forget to include the length and the width in your answer.
The length of a rectangle is 2 more than 3 times the width , if the perimeter is 100 meters what is the width of the rectangle
To solve for x and find the dimensions of the rectangle, we can follow these steps:
Step 1: Set up the equation based on the given information.
We are told that the length of the rectangle is 1 cm longer than its width. So, let's define:
x = width of the rectangle
x + 1 = length of the rectangle
Step 2: Apply the Pythagorean theorem.
According to the Pythagorean theorem, in a right-angled triangle, the square of the length of the hypotenuse (the diagonal) is equal to the sum of the squares of the other two sides. In our case, the width and the length of the rectangle form two sides of the right-angled triangle, and the diagonal is the hypotenuse.
The Pythagorean theorem equation is:
a^2 + b^2 = c^2
Substituting the values from the rectangle, we have:
x^2 + (x + 1)^2 = 4^2
Step 3: Simplify the equation.
Expand the square on the left side of the equation:
x^2 + (x^2 + 2x + 1) = 16
Combine like terms:
2x^2 + 2x + 1 = 16
Step 4: Move all terms to one side of the equation to solve for x.
Rearrange the equation by subtracting 16 from both sides:
2x^2 + 2x + 1 - 16 = 0
Simplify:
2x^2 + 2x - 15 = 0
Step 5: Solve the quadratic equation.
To solve the quadratic equation, we can use factoring, quadratic formula, or completing the square. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = 2, and c = -15. Substituting these values into the quadratic formula:
x = (-2 ± √(2^2 - 4(2)(-15))) / (2(2))
Simplify the expression under the square root:
x = (-2 ± √(4 + 120)) / 4
x = (-2 ± √124) / 4
x = (-2 ± 2√31) / 4
Step 6: Solve for x.
Now we have two possible solutions for x:
x = (-2 + 2√31) / 4
x = (-2 - 2√31) / 4
Simplify each expression:
x = -1 + √31 / 2
x = -1 - √31 / 2
Step 7: Calculate the length and width.
Since we defined x as the width of the rectangle, substituting the values back into our original definitions:
Length = x + 1
Using the first solution for x:
Length = -1 + √31 / 2 + 1
Length = √31 / 2
Width = -1 + √31 / 2
Using the second solution for x:
Length = -1 - √31 / 2 + 1
Length = -1 - √31 / 2
Width = -1 - √31 / 2
Note: While the width cannot be negative in this context, it is important to include both solutions since the quadratic equation can have both positive and negative roots. In this case, the negative root is not applicable to the dimensions of the rectangle.
Therefore, the dimensions of the rectangle (to the nearest thousandth) are:
Length ≈ √31 / 2 cm
Width ≈ √31 / 2 cm