Question

what is the most precise name for quadrilateral abcd with vertices a(-5, -1) b(-5, 3) c(-2, 3) d(-2, -1)

Answer 1

check the picture below.

is a parallelogram, it has two pairs of sides that are parallel.

hmmm is not a square, since is a 3x4, so all sides aren't equal.

is a quadrilateral, it has four sides alright.

though is not a square, it does have 4 right-angles at its corners... that means that is a rectangle, besides being a parallelogram.

Answer 2

Answer:

1st Option is correct.

Step-by-step explanation:

Given:

Vertices of the quadrilateral ABCD.

A( -5 , -1 ) , B( -5 , 3 ) , C( -2 , 3 ) , D( -2 , -1 )

To find: Name of the Quadrilateral.

We use Distance formula to find the length of the sides and diagonal of the Quadrilateral.

Distance between two point = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Length of Side AB $=\sqrt{(-5-(-5))^2+(-1-3)^2}=\sqrt{(0)^2+(-4)^2}=\sqrt{0+16}=4$

Length of Side CB $=\sqrt{(-2-(-5))^2+(3-3)^2}=\sqrt{(-2+5)^2+(0)^2}=\sqrt{9+0}=3$

Length of Side CD $=\sqrt{(-2-(-2))^2+(-1-3)^2}=\sqrt{(0)^2+(-4)^2}=\sqrt{0+16}=4$

Length of Side AD $=\sqrt{(-2-(-5))^2+(-1-(-1))^2}=\sqrt{(3)^2+(0)^2}=\sqrt{9+0}=3$

Length of the Diagonal AC $=\sqrt{(-2-(-5))^2+(3-(-1))^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=5$

Length of the Diagonal BD $=\sqrt{(-2-(-5))^2+(-1-3)^2}=\sqrt{(3)^2+(-4)^2}=\sqrt{9+16}=5$

So, Opposite side of the Quadrilateral are Equal that is AB = CD = 4 unit and CB = AD = 3 unit

Also, Diagonals are equal that is AC = BD = 5 unit

⇒  Quadrilateral is a RECTANGLE.

Therefore, 1st Option is correct.