Find the area of rectangle ABCD with vertices A(-3, 0), B(3, 2), C(4, -1), and D(-2, -3).
AB = √((3+3)^2 + 2^2) = √40
BC = √((4-3)^2 + (-1-2)^2) = √10
So the area = length * width = √40 * √10 = √400 = 20
I'm assuming that point D is correctly placed, and that AB⊥BC, since they said it was a rectangle.
I have no idea
To find the area of a rectangle, you need two adjacent sides of the rectangle. In this case, we have the coordinates of the four vertices A(-3, 0), B(3, 2), C(4, -1), and D(-2, -3).
First, let's find the length of one side of the rectangle. We can use the distance formula to find the length between two adjacent vertices A and B:
Distance AB = √[(x₂ - x₁)² + (y₂ - y₁)²]
x₁ = -3, y₁ = 0 (coordinates of A)
x₂ = 3, y₂ = 2 (coordinates of B)
Plugging in the values, we get:
Distance AB = √[(3 - (-3))² + (2 - 0)²]
= √[6² + 2²]
= √[36 + 4]
= √40
= 2√10
Now, let's find the length of the other side of the rectangle. We can use the distance formula again to find the length between vertices B and C:
Distance BC = √[(x₂ - x₁)² + (y₂ - y₁)²]
x₁ = 3, y₁ = 2 (coordinates of B)
x₂ = 4, y₂ = -1 (coordinates of C)
Plugging in the values, we get:
Distance BC = √[(4 - 3)² + (-1 - 2)²]
= √[1² + (-3)²]
= √[1 + 9]
= √10
Now that we have the length of two adjacent sides of the rectangle, we can find the area:
Area of rectangle = length × width
Length = 2√10 (distance AB)
Width = √10 (distance BC)
Area of rectangle = 2√10 × √10
= 2√10 × √10
= 2 × 10
= 20
Therefore, the area of rectangle ABCD is 20 square units.