+274 In the diagram, four squares of side length 2 are placed in the corners of a square of side length 6. Each of the points W, X, Y, and Z is a vertex of one of the small squares. Square ABCD can be constructed with sides passing through W, X, Y, and Z. What is the maximum possible distance from A to P?https://latex.artofproblemsolving.com/3/a/e/3ae9131cc46a1aa7c3ba6923b1dfde6b8cee4537.png I would like a solution preferably instead of an answer because I'm trying to prepare for a final.
+274 Nope, can you please help? I'll be active again in couple of hours. I need to take a break from math; I have been doing math for 3 hours now.
To find the maximum possible distance from point A to point P in the given diagram, we need to determine the position of point P relative to the square ABCD.
Let's analyze the diagram step by step:
The large square ABCD has a side length of 6 units.
Four smaller squares are placed in the corners of ABCD, with side length 2 units each.
The vertices of these smaller squares are labeled as W, X, Y, and Z.
We want to find the maximum distance from point A to point P.
To maximize the distance from A to P, we should consider the scenario where P is located at the opposite corner of the large square ABCD relative to A. In this case, P would be at point D, which is the farthest possible point from A.
Therefore, the maximum possible distance from A to P is equal to the diagonal of the large square ABCD. Using the Pythagorean theorem, we can calculate this diagonal length:
Diagonal = √(Side^2 + Side^2) = √(6^2 + 6^2) = √(36 + 36) = sqrt(72) = 6*sqrt(2).
