Question

Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-6, -4),A(−6,−4),A, left parenthesis, minus, 6, comma, minus, 4, right parenthesis, comma B(-4,-4)B(−4,−4)B, left parenthesis, minus, 4, comma, minus, 4, right parenthesis, C(-4, -2)C(−4,−2)C, left parenthesis, minus, 4, comma, minus, 2, right parenthesis, and D(-6, -2)D(−6,−2)D, left parenthesis, minus, 6, comma, minus, 2, right parenthesis.
What is the perimeter of rectangle ABCDABCDA, B, C, D?

Answer 1

Answer with explanation:

The coordinates of vertices of Rectangle ABCD are ,A(-6,-4),B(-4,-4),C(-4,-2) and (-6,-2).

Distance formula ,for finding the distance between two points ,(a,b) and (c,d)

     $=\sqrt{(a-c)^2+(b-d)^2}$

  $AB=CD=\sqrt{(-6+4)^2+(-4+4)^2}=\sqrt{2^2}=2\\\\BC=AD=\sqrt{(-4+4)^2+(-4+2)^2}=\sqrt{2^2}=2$

Perimeter of Rectangle ,ABCD , which is a square

=Sum of Length of all sides

  = AB + BC+CD+DA

    = 4 × Length of any side

   = 2 +2+2+2 or → 4 × 2

   = 8 units  

Answer 2

Answer:

8 Units

Step-by-step explanation:

A (-6 , -4)

B (-4 , -4)

C (-4 , -2)

D (-6 , -2)

AB = √( (-4-(-6))² + (-4-(-4))²) = √ (2² + 0²) = √(4 + 0) = 2

BC = √( (-4-(-4))² + (-2-(-4))²) = √ (0² + 2²) = √(0 + 4) = 2

CD = √( (-6-(-4))² + (-2-(-2))²) = √ ((-2)² + 0²) = √(4 + 0) = 2

AD = √( (-6-(-6))² + (-2-(-4))²) = √ (0² + 2²) = √(0 + 4) = 2

Perimeter of ABCD = AB + BC + CD + DA

= 2 + 2 + 2 + 2

= 8 Units