How to find the area of a square abcd in looking for pythagoras?

A = 25 units²

Step-by-step explanation:

You need to visualize two triangles on the lower left and right that are OUTSIDE of the square.

One leg of the first triangle is formed by starting at Point A and going down 4 units.  The other leg is formed by starting at point D and moving left 3 units.  Line segment AD is the hypotenuse of this triangle.  

The leg lengths are 3 and 4, so the hypotenuse is...

3² + 4² = c²

   9 + 16 = c²

      25 = c²

          5 = c

So side AD of the square is 5 units long.  

We can do the same concept using points C and D.  We can make a triangle that has legs with length 3 and 4 with CD being the hypotenuse.  The length of CD is therefore 5 as well.

The area of the square is

A = 5*5 = 25 units²

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .