To find the length and width of a rectangle, we can use the two given pieces of information: the perimeter and the area.

Step 1: Define the variables

Let:

• $L$ = length of the rectangle
• $W$ = width of the rectangle

Step 2: Use the formulas for perimeter and area

1. Perimeter formula: $P = 2L + 2W$ We're given that the perimeter $P = 42$ feet, so: $2L + 2W = 42$ Dividing both sides by 2: $L + W = 21 \quad \text{(Equation 1)}$

2. Area formula: $A = L \times W$ We're given that the area $A = 90$ square feet, so: $L \times W = 90 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations

We have the system of two equations:

$L + W = 21 \quad \text{(Equation 1)}$ $L \times W = 90 \quad \text{(Equation 2)}$

Solve for one variable

From Equation 1, solve for $L$ in terms of $W$: $L = 21 - W$

Substitute into the area equation

Substitute $L = 21 - W$ into Equation 2: $(21 - W) \times W = 90$ Expanding this: $21W - W^2 = 90$ Rearranging the terms: $W^2 - 21W + 90 = 0$

Solve the quadratic equation

Now, solve the quadratic equation $W^2 - 21W + 90 = 0$. Using the quadratic formula: $W = \frac{-(-21) \pm \sqrt{(-21)^2 - 4(1)(90)}}{2(1)}$ $W = \frac{21 \pm \sqrt{441 - 360}}{2}$ $W = \frac{21 \pm \sqrt{81}}{2}$ $W = \frac{21 \pm 9}{2}$ So, we have two solutions for $W$: $W = \frac{21 + 9}{2} = 15 \quad \text{or} \quad W = \frac{21 - 9}{2} = 6$

Find $L$

For $W = 15$, substitute into $L = 21 - W$: $L = 21 - 15 = 6$

For $W = 6$, substitute into $L = 21 - W$: $L = 21 - 6 = 15$

Step 4: Conclusion

The length and width of the rectangle are 15 feet and 6 feet, respectively. The order of length and width does not matter as they are interchangeable in this case.

Thus, the dimensions of the rectangle are:

• Length = 15 feet
• Width = 6 feet

Would you like further details on the steps or have any questions about this process?

Here are some related questions you may find interesting:

1. How do you derive the quadratic equation from a real-world problem?
2. How do changes in perimeter and area affect the dimensions of a rectangle?
3. What other shapes use similar methods for calculating dimensions based on area and perimeter?
4. How does the concept of optimization relate to finding dimensions for maximal or minimal area?
5. What is the general method for solving systems of equations when both equations are quadratic?

Tip: If you're ever stuck on a quadratic equation, always check if it can be factored before resorting to the quadratic formula!