https://file.aiquickdraw.com/mathbot/file/1728956342528l825k4n9.jpg

Let's analyze the question.

We are given that the diagonals of quadrilateral \(WXYZ\) are **perpendicular bisectors** of each other, meaning:
1. The diagonals intersect at right angles (90°) at point \(P\).
2. Each diagonal is bisected at the point of intersection.

### Known values:
- \(WP = 24 \, \text{cm}\), so the entire diagonal \(WZ\) is \(2 \times 24 = 48 \, \text{cm}\).
- \(XP = 7 \, \text{cm}\), so the entire diagonal \(XY\) is \(2 \times 7 = 14 \, \text{cm}\).

Since the diagonals are perpendicular bisectors of each other, the quadrilateral \(WXYZ\) is a **rhombus** (a special kind of parallelogram). 

### Finding the side length of the rhombus:
We can use the **Pythagorean theorem** to find the length of each side of the rhombus. Each side (e.g., \(WX\)) forms a right triangle with half of the diagonals:
\[
WX^2 = WP^2 + XP^2
\]
\[
WX^2 = 24^2 + 7^2 = 576 + 49 = 625
\]
\[
WX = \sqrt{625} = 25 \, \text{cm}
\]

### Perimeter of the rhombus:
Since all sides of a rhombus are equal, the perimeter is:
\[
\text{Perimeter} = 4 \times WX = 4 \times 25 = 100 \, \text{cm}
\]

Thus, the perimeter of \(WXYZ\) is:
\[
\boxed{100 \, \text{cm}}
\]

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The diagonals of quadrilateral WXYZ are perpendicular bisectors of each other, and intersect at point P. WP is 24 cm long, and XP is 7 cm long. What is the perimeter of WXYZ?

Learn how to find the perimeter of quadrilateral WXYZ, given its diagonals are perpendicular bisectors and intersect at a point. This problem covers key geometry concepts, including the Pythagorean theorem and properties of rhombuses, suitable for students in Grades 8-10.

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