Question

The coordinates of three vertices of square ABCD are A(−212,112),B(−212,−3), and C(2,112).
When point D is placed on this square, what will the perimeter of the square be?

Answer 1

Given info : The coordinates of three vertices of rectangle (not square) ABCD are A(−212,112) , B(−212,−3) , and C(2,112).

To find : point D and perimeter of rectangle.

Solution : let's locate the point on Cartesian system. See figure here it has clearly shown that coordinate of point D = (2, 3)

If you didn't get by Cartesian system.

You can find it by analytical method.

You know, midpoints of diagonals of rectangle ( a parallelogram) must be met at the same point.

I.e., midpoint of AD = midpoint of BC

⇒[ (-212 + x)/2 , (112 + y)/2] = [(-212+2)/2 , (112 - 3)/2]

You will get, D = (2, -3).

Now let's find perimeter of rectangle = 2(AB + BD)

You can find by anyone of them.

AB = √{(-212 + 212)² + (112 + 3)²} = 115

BD = √{(-212 - 2)² + (-3 + 3)²} = 214

Now, perimeter of rectangle ABCD = 2(115 + 214) = 2(329) = 658 unit

Answer 2

Answer:it is literally just  658

Step-by-step explanation: