Consider a shape with vertices A(1, 4), B(3, 0), C(1, −4), and D(−1, 0) on the coordinate plane. 1) Which proves that the shape given by the vertices is a rhombus? A) AB = BC = CD = DA = 10 B) AB = BC = CD = DA = 15 C) AB = BC = CD = DA = 2 5 D) AB = BC = CD = DA = 3 5

C) AB = BC = CD = DA = 2√5

Step-by-step explanation:

Given

A(1, 4)

B(3, 0)

C(1, −4)

D(−1, 0)

We get the distance between point A and point B

AB = √((3-1)²+(0-4)²) = √20 = 2√5

BC = √((1-3)²+(-4-0)²) = √20 = 2√5

CD = √((-1-1)²+(0-(-4))²) = √20 = 2√5

DA = √((1-(-1))²+(4-0)²) = √20 = 2√5

C) AB = BC = CD = DA = 2√5

C) AB = BC = CD = DA = 2√5

Corrected question:

Consider a shape with vertices A(1, 4), B(3, 0), C(1, −4), and D(−1, 0) on the coordinate plane. 1) Which proves that the shape given by the vertices is a rhombus? A) AB = BC = CD = DA = 10 B) AB = BC = CD = DA = 15 C) AB = BC = CD = DA = 2√5 D) AB = BC = CD = DA = 3√5

Step-by-step explanation:

Given;

Vertices

A(1,4)

B(3,0)

C(1,-4)

D(-1,0)

We need to determine the Length of sides;

AB,BC,CD,DA

Length = √((∆x)^2 + (∆y)^2)

For

AB = √((3-1)^2 + (0-4)^2) = √(4+16) = √20 = 2√5

BC = √((1-3)^2 + (-4-0)^2) = √(4+16) = √20 = 2√5

CD = √((-1-1)^2 + (0--4)^2) = √(4+16) = √20 = 2√5

DA = √((1--1)^2 + (4-0)^2) = √(4+16) = √20 = 2√5

Which shows that;

AB=BC=CD=DA=2√5

For a rhombus, all sides are equal.

Therefore, AB=BC=CD=DA=2√5, proves that the shape given by the vertices is a rhombus.

idc

step-by-step explanation:

idk