Question

HELLLLLLP PLEASE CAN SOMEONE PLEASE EXPLAIN?
Consider parallelogram ABCD with vertices A(-6, 6), B(-2, 8), C(0, 4), and D(-4, 2). Classify the parallelogram and select ALL that apply.
Group of answer choices

ABCD is a rectangle.

ABCD is none of these.

ABCD is a rhombus.

ABCD is a square.

Answer 1

Answer:

ABCD is a RHOMBUS or a SQUARE.

Step-by-step explanation:

The coordinates are A(-6, 6), B(-2, 8), C(0, 4), and D(-4, 2).

By DISTANCE FORMULA:

The length of the segment with coordinates X(a,b) and Y(c,d) is given as:

$XY = \sqrt{(c-a)^2  + (d-b)^2}$

Now, similarly, the lengths of the segments are:

$AB = \sqrt{(-2 -(-6))^2  + (8-6)^2}$

$=\sqrt{(4)^2  + (2)^2}  = \sqrt{16 + 4}   = \sqrt{20}$

⇒ The length of the segment AB = √ 20 units

$BC = \sqrt{(0 -(-2))^2  + (4-8)^2}$

$=\sqrt{(2)^2  + (-4)^2}  = \sqrt{4 + 16}   = \sqrt{20}$

⇒ The length of the segment BC = √ 20 units

$CD = \sqrt{(0 -(-4))^2  + (2-4)^2}$

$=\sqrt{(4)^2  + (-2)^2}  = \sqrt{16 + 4}   = \sqrt{20}$

⇒ The length of the segment CD= √ 20 units

$AD = \sqrt{(-6 -(-4))^2  + (6-2)^2}$

$=\sqrt{(-2)^2  + (4)^2}  = \sqrt{4 + 16}   = \sqrt{20}$

⇒ The length of the segment AD = √ 20 units

$AC = \sqrt{(0 -(-6))^2  + (4-6)^2}$

$=\sqrt{(6)^2  + (-2)^2}  = \sqrt{36 + 4}   = \sqrt{40}$

⇒ The length of the diagonal AC = √ 40 units

$BD = \sqrt{(-2 +4)^2  + (8-2)^2}$

$=\sqrt{(2)^2  + (6)^2}  = \sqrt{36 + 4}   = \sqrt{40}$

⇒ The length of the diagonal BD = √ 40 units

Since, here the length of all segments is √ 20 units.

⇒AB = BC = CD = AD = √ 20 units

and Diagonal AC = BD

⇒ ABCD is a RHOMBUS or SQUARE.