MATH SOLVE

9 months ago

Q:
# Quadrilateral JKLM has vertex coordinates J(2,4), K(6,1), L(2,-2), and M(-2,1). What type of quadrilateral is JKLM?

Accepted Solution

A:

Find lengths of quadrilateral sides:

[tex] |JK|=\sqrt{(6-2)^2+(1-4)^2} =\sqrt{16+9}=5 [/tex],

[tex] |KL|=\sqrt{(2-6)^2+(-2-1)^2} =\sqrt{16+9}=5 [/tex],

[tex] |LM|=\sqrt{(2-(-2))^2+(-2-1)^2} =\sqrt{16+9}=5 [/tex],

[tex] |MJ|=\sqrt{(-2-2)^2+(1-4)^2} =\sqrt{16+9}=5 [/tex].

Since all sides have the same lengths, you can state that this quadrilateral is rhombus.

Let's check whether quadrilateral JKLM is a square. To check this let find the lengths of diagonals:

[tex] |JL|=\sqrt{(2-2)^2+(-2-4)^2} =\sqrt{0+36}=6 [/tex], [tex] |MK|=\sqrt{(-2-6)^2+(1-1)^2} =\sqrt{64+0}=8 [/tex].

The lengths are different, so quadrilateral is not a square.

Answer: quadrilateral JKLM is a rhombus.

[tex] |JK|=\sqrt{(6-2)^2+(1-4)^2} =\sqrt{16+9}=5 [/tex],

[tex] |KL|=\sqrt{(2-6)^2+(-2-1)^2} =\sqrt{16+9}=5 [/tex],

[tex] |LM|=\sqrt{(2-(-2))^2+(-2-1)^2} =\sqrt{16+9}=5 [/tex],

[tex] |MJ|=\sqrt{(-2-2)^2+(1-4)^2} =\sqrt{16+9}=5 [/tex].

Since all sides have the same lengths, you can state that this quadrilateral is rhombus.

Let's check whether quadrilateral JKLM is a square. To check this let find the lengths of diagonals:

[tex] |JL|=\sqrt{(2-2)^2+(-2-4)^2} =\sqrt{0+36}=6 [/tex], [tex] |MK|=\sqrt{(-2-6)^2+(1-1)^2} =\sqrt{64+0}=8 [/tex].

The lengths are different, so quadrilateral is not a square.

Answer: quadrilateral JKLM is a rhombus.