SOLUTION: Square ABCD has vertices A(1,—2) and C(4, 5). What is the area of the square?

Algebra ->  Length-and-distance -> SOLUTION: Square ABCD has vertices A(1,—2) and C(4, 5). What is the area of the square?      Log On


   

Question 1178917: Square ABCD has vertices A(1,—2) and C(4, 5). What
is the area of the square?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Square ABCD+has vertices A(1,-2) and C(4,+5).
the length of the side is equal to distance between vertices
AC=a
a=sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29
a=sqrt%28%284-1%29%5E2%2B%285-%28-2%29%29%5E2%29
a=sqrt%283%5E2%2B%285%2B2%29%5E2%29
a=sqrt%283%5E2%2B7%5E2%29
a=sqrt%289%2B49%29
a=sqrt%2858%29


then the area of the square is
A=a%5E2
A=%28sqrt%2858%29%29%5E2
A=58


Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by  @MathLover1 is  INCORRECT.

            It is incorrect,  because she mistakenly treats  AC  as a side of the square ABCD,
            while it is   THE  DIAGONAL   of the square.

            I came to bring the correct solution. 


|AC| = sqrt%28%284-1%29%5E2+%2B+%285-%28-2%29%29%5E2%29 = sqrt%283%5E2+%2B+7%5E2%29 = sqrt%2858%29.


The area of a square is HALF THE PRODUCT of its diagonals,

or, equivalently, HALF of the SQUARE of the diagonal


    AREA%5BABCD%5D = %281%2F2%29%2A%28abs%28AC%29%29%5E2 = %281%2F2%29%2A%28sqrt%2858%29%29%5E2%29 = %281%2F2%29%2A58 = 29 square units.    ANSWER

Solved (correctly)

----------------

For your safety,  ignore the post by  @MathLover1.