In order to find the Perimeter of the Rectangle, First we need to find the Length and Width of the Rectangle.
Distance between two points (x₁ , y₁) and (x₂ , y₂) is given by :
[tex]\bigstar\;\; \mathsf{Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}[/tex]
There are two lengths in a rectangle. Let us find any one length of the rectangle.
I'm considering the length with co-ordinates (-8 , 2) and (-2 , 10)
Here : x₁ = -8 and x₂ = -2 and y₁ = 2 and y₂ = 10
Substituting the values in the distance formula, We get :
[tex]\mathsf{\implies Length = \sqrt{[-2 - (-8)]^2 + [10 - 2]^2}}[/tex]
[tex]\mathsf{\implies Length = \sqrt{[-2 + 8]^2 + [8]^2}}[/tex]
[tex]\mathsf{\implies Length = \sqrt{[6]^2 + [8]^2}}[/tex]
[tex]\mathsf{\implies Length = \sqrt{36 + 64}}[/tex]
[tex]\mathsf{\implies Length = \sqrt{100}}[/tex]
[tex]\mathsf{\implies Length = 10}[/tex]
There are two widths in a rectangle. Let us find any one width of the rectangle.
I'm considering the width with co-ordinates (-2 , 10) and (2 , 7)
Here : x₁ = -2 and x₂ = 2 and y₁ = 10 and y₂ = 7
Substituting the values in the distance formula, We get :
[tex]\mathsf{\implies Width = \sqrt{[2 - (-2)]^2 + [7 - 10]^2}}[/tex]
[tex]\mathsf{\implies Width = \sqrt{[2 + 2]^2 + [-3]^2}}[/tex]
[tex]\mathsf{\implies Width = \sqrt{[4]^2 + [-3]^2}}[/tex]
[tex]\mathsf{\implies Width = \sqrt{16 + 9}}[/tex]
[tex]\mathsf{\implies Width = \sqrt{25}}[/tex]
[tex]\mathsf{\implies Width = 5}[/tex]
Perimeter of a Rectangle is given by : 2[Length + Width]
[tex]:\implies[/tex] Perimeter of the given rectangle = 2[10 + 5]
[tex]:\implies[/tex] Perimeter of the given rectangle = 2[15]
[tex]:\implies[/tex] Perimeter of the given rectangle = 30
Answer : Perimeter of the given rectangle is 30 square units
Answer:
A
Step-by-step explanation:
The opposite sides of a rectangle are congruent.
Calculate the lengths of 2 of the sides using the distance formula and multiply by 2
d = √ (x₂ - x₁ )² + (y₂ - y₁ )²
with (x₁, y₁ ) = (- 2, 10) and (x₂, y₂ ) = (2, 7)
d = [tex]\sqrt{(2+2)^2+(7-10)^2}[/tex]
= [tex]\sqrt{4^2+(-3)^2}[/tex]
= [tex]\sqrt{16+9}[/tex] = [tex]\sqrt{25}[/tex] = 5
Repeat
with (x₁, y₁ ) = (- 8, 2) and (x₂, y₂ ) = (- 2, 10)
d = [tex]\sqrt{(-2+8)^2+(10-2)^2}[/tex]
= [tex]\sqrt{6^2+8^2}[/tex]
= [tex]\sqrt{36+64}[/tex] = [tex]\sqrt{100}[/tex] = 10
Hence
Perimeter = (2 × 5) + (2 × 10) = 10 + 20 = 30 units → A
