To find the length of one of the diagonals of the rectangle, we can use the information provided: the area and perimeter.

Let's denote the length and width of the rectangle as $l$ and $w$, respectively.

Step 1: Set up the equations using the given information.

• Area: $l \times w = 68$
• Perimeter: $2l + 2w = 71$

Step 2: Express one variable in terms of the other using the perimeter equation.

$2l + 2w = 71 \quad \Rightarrow \quad l + w = \frac{71}{2} = 35.5$ Thus, $l = 35.5 - w$.

Step 3: Substitute $l = 35.5 - w$ into the area equation.

$(35.5 - w) \times w = 68$ This simplifies to: $35.5w - w^2 = 68$ $w^2 - 35.5w + 68 = 0$

Step 4: Solve the quadratic equation.

To solve the quadratic equation $w^2 - 35.5w + 68 = 0$, we can use the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -35.5$, and $c = 68$.

$w = \frac{35.5 \pm \sqrt{(35.5)^2 - 4 \times 1 \times 68}}{2 \times 1}$ $w = \frac{35.5 \pm \sqrt{1260.25 - 272}}{2}$ $w = \frac{35.5 \pm \sqrt{988.25}}{2}$ $w = \frac{35.5 \pm 31.43}{2}$ This gives us two possible values for $w$: $w_1 = \frac{35.5 + 31.43}{2} = 33.465 \quad \text{or} \quad w_2 = \frac{35.5 - 31.43}{2} = 2.035$ Since the larger value is likely the length and the smaller one is the width, we can take $w_2$ as $w = 2.035$ and $l = 33.465$.

Step 5: Find the length of the diagonal.

The length of the diagonal $d$ can be found using the Pythagorean theorem: $d = \sqrt{l^2 + w^2}$ $d = \sqrt{33.465^2 + 2.035^2} \approx \sqrt{1120.87 + 4.14} \approx \sqrt{1125.01} \approx 33.55$

So, the length of the diagonal is approximately 33.55 units.

Would you like to go over any particular step or need further explanation?

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Here are some related questions for further practice:

1. How would you solve for the length of the diagonal if only the area was given?
2. What happens to the diagonal length if the area is doubled but the perimeter remains the same?
3. Can you derive a general formula for the diagonal of a rectangle in terms of its area and perimeter?
4. How do changes in the ratio of length to width affect the diagonal length?
5. What is the significance of the diagonal in real-world applications like construction or design?

Tip: When dealing with quadratic equations, always check if the discriminant ($b^2 - 4ac$) is positive to ensure real solutions.