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The problem states that the perimeter of quadrilateral \( PQRS \) is 84, and we need to find the length of each side under the given conditions.

For problem 14, the given condition is \( PQ = QR \). 

### Steps to solve:
1. Since the perimeter is 84, and the perimeter is the sum of the lengths of all sides, we have the equation:
   \[
   PQ + QR + RS + SP = 84
   \]
2. The condition \( PQ = QR \) simplifies this equation:
   \[
   2PQ + RS + SP = 84
   \]
3. Since no further information is provided about \( RS \) and \( SP \), it's common to assume that the quadrilateral is a rectangle or square. If this is the case, we can assume \( RS = SP \).

4. Let’s denote \( PQ = QR = x \) and \( RS = SP = y \). This gives the equation:
   \[
   2x + 2y = 84
   \]
   Simplifying:
   \[
   x + y = 42
   \]
5. Without more specific conditions, \( x \) and \( y \) must be positive values that satisfy \( x + y = 42 \).

If you are looking for specific integer solutions, you can choose any pair \( x \) and \( y \) such that their sum is 42. If you need more clarity, feel free to provide additional conditions or assumptions (e.g., if it is a square or rectangle).

Do you have any questions, or would you like further details?

Here are some related questions to extend the understanding:
1. What would happen if \( PQ \) and \( QR \) were not equal?
2. How would you solve this problem if \( PQRS \) was a rectangle?
3. What would change if \( PQRS \) was a square?
4. How can you find the area of \( PQRS \) if you know the side lengths?
5. How does the solution change if one of the sides is doubled in length?

**Tip:** Always check for symmetry in geometric problems, as it often simplifies the problem-solving process!

The perimeter of quadrilateral PQRS is 84. Find the length of each side of PQRS under the given conditions where PQ = QR.

Explore how to determine the side lengths of a quadrilateral when two adjacent sides are equal and the total perimeter is given. This problem involves using algebraic methods to solve for sides, suitable for students in Grades 8-10.

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