Question

The area of square ABCD with vertices at A(1,-1), B(5,-3), C(7,1), and D(3,3) isapproximately 20.25 square units.True or False

Answer 1

Given:

A(1,-1), B(5,-3), C(7,1), and D(3,3)

Area of ABCD = 20.25 square units

Let's determine if the area is correct.

Since ABCD is a square, all side lengths are equal.

Now, let's find the length of one side.

Apply the distance formula:

`√((x2-x1)^2+(y2-y1)^2)`

Let's find the length of AB.

Where:

(x1, y1) ==> A(1, -1)

(x2, y2) ==> B(5, -3)

Thus, we have:

`\begin{gathered} AB=√((5-1)^2+(-3-(-1))^2) \\ \\ AB=√(4^2+(-3+1)^2) \\ \\ AB=√(16+(-2)^2) \\ \\ AB=√(16+4) \\ \\ AB=√(20) \end{gathered}`

The length of one side of the square is √20.

Now, to find the area of a square, we have:

`\begin{gathered} Area=l^2 \\ \\ Area=(√(20))^2 \\ \\ Area=20\text{ square units} \end{gathered}`

Therefore, the area of the square is 20 square units.

This means the area is not 20.25 square units.

Therefore, the