The lengths of the four sides of a quadrilateral (in inches) are consecutive even integers. If the perimeter is 5252 inches, find the value of the longest of the four side lengths.

Let's denote the lengths of the four sides of the quadrilateral as consecutive even integers. We can express them as follows:

- First side: \( x \) (the smallest even integer)
- Second side: \( x + 2 \)
- Third side: \( x + 4 \)
- Fourth side: \( x + 6 \)

The perimeter of the quadrilateral is the sum of all its side lengths. Therefore, the perimeter is given by:

\[
x + (x + 2) + (x + 4) + (x + 6) = 4x + 12
\]

According to the problem, the perimeter is 5252 inches. So, we can set up the equation:

\[
4x + 12 = 5252
\]

Now, let's solve for \( x \):

\[
4x = 5252 - 12
\]
\[
4x = 5240
\]
\[
x = \frac{5240}{4} = 1310
\]

So, the smallest side length is 1310 inches. The consecutive even integers are:

- First side: \( 1310 \)
- Second side: \( 1312 \)
- Third side: \( 1314 \)
- Fourth side: \( 1316 \)

The longest side is \( 1316 \) inches.

### Final answer:
The longest side length is \( 1316 \) inches.

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the problem change if the integers were consecutive odd numbers instead?
2. Can the formula for the perimeter of a quadrilateral change depending on its shape?
3. What is the sum of the squares of the four side lengths?
4. How can this method be extended to solve for the sides of a polygon with more sides?
5. What if the perimeter was a different value, say 6000 inches?

**Tip:** Consecutive even integers can be modeled by an arithmetic sequence with a common difference of 2, making it easy to express their sum algebraically!

Learn how to find the longest side of a quadrilateral where the sides are consecutive even integers, and the perimeter is 5252 inches. This algebra problem involves calculating the sum of consecutive even integers and solving for the longest side length, suitable for students in Grades 6-8.

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