Answer:
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.
