Find the perimeter of the following shape, rounded to the nearest tenth: Shape ABCD is shown. Point A is at negative 3, 5. Point B is at 1, 7. Point C is at 3, 3. Point D is at negative 1, 1. 12 13.9 17.9 20

Question:

Find the perimeter of the following shape, rounded to the nearest tenth: Shape ABCD is shown. Point A is at negative 3, 5. Point B is at 1, 7. Point C is at 3, 3. Point D is at negative 1, 1.

a) 12

b) 13.9

c) 17.9

d) 20

c) 17.9

Step-by-step explanation:

In the question, we are given the following points

Point A is at negative 3, 5. = (-3, 5)

Point B is at 1, 7 =  (1, 7)

Point C is at 3, 3 = (3, 3)

Point D is at negative 1, 1 = (-1, 1)

When we are given points (x₁, y₁)  and ( x₂ , y₂) we would use the formula

√(x₂ - x₁)² + (y₂ - y₁)²

For point A and Point B = AB

= √(x₂ - x₁)²+ (y₂ - y₁)²

= √(1 - (-3) )² + (7 - 5)²

= √(4² + 2²)

= √(16 + 4)

= √20

For point B and Point C = BC

= √(x₂ - x₁)²+ (y₂ - y₁)²

= √(3 - 1 )² + (3 - 7)²

= √(2² + (-4)²)

= √4 + 16

= √20

For point D and Point C = DC

= √(x₂ - x₁)²+ (y₂ - y₁)²

= √(-1 - (-3) )² + (1 - 3)²

= √(4² + -2²)

= √(16 + 4)

= √20

For point A and Point D = AD

= √(x₂ - x₁)²+ (y₂ - y₁)²

= √(-1 - (-3) )² + (1 - 5)²

= √(-2² + -4²)

= √(4 + 16)

= √20

From the solved calculations above, we can see that

AB = BC = DC = AD

This proves to us, that the above shape with points A, B, C and D is a square.

This is because the sides of a square are equal to one another

The formula for the perimeter of a square = 4 × Length of one side

Hence, the perimeter of this square in the above question =

4 × √20

17.88854382.

Approximately to the nearest tenth = 17.9

Therefore, the perimeter of the following shape ( Square) to the nearest tenth = 17.9

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.