Question

Parallelogram ABCD has vertices at A(0, 0) , B(4, 5) , C(8, 0) , and D(4,−5) .

Which conclusion can be made?


1. AB¯¯¯¯¯⊥BC¯¯¯¯¯ ; therefore, ABCD is a square.

2. AB¯¯¯¯¯⊥BC¯¯¯¯¯ ; therefore, ABCD is a rectangle.

3. AB=BC ; therefore ABCD is a square.

4. AB=BC ; therefore, ABCD is a rhombus.

Answer 1

AB = √(5-0)^2 + (4-0)^2 = √25+16 = √41

BC = √(0-5)^2 + (8-4)^2 = √25+16 = √41

CD = √(-5 - 0)^2 + (4-8)^2 = √25+16 = √41

AD = √(-5-0)^2 + (4-0)^2 = √25+16 = √41

All the sides are equal so it is either a square or a rhombus.

The distance from A to C on the x axis is: 8-0 = 8 units.

The distance from B to D on the Y axis is: 5 - - 5 = 10 units.

Because the two distances are not the same, it is not a square.

The answer is 4. AB=BC ; therefore, ABCD is a rhombus.