Question

Pleasee help ASAP!! No links pleaseee!!!

Answer 1

Given:

The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).

To find:

The area of the rectangle.

Solution:

Distance formula:

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

Using the distance formula, we get

`AB=√((2-0)^2+(4-1)^2)`

`AB=√((2)^2+(3)^2)`

`AB=√(4+9)`

`AB=√(13)`

Similarly,

`BC=√((6-2)^2+(0-4)^2)`

`BC=√((4)^2+(-4)^2)`

`BC=√(16+16)`

`BC=√(32)`

`BC=4√(2)`

Now, the length of the rectangle is
`AB=√(13)` and the width of the rectangle is
`BC=4√(2)`. So, the area of the rectangle is:

`A=length * width`

`A=√(13)* 4√(2)`

`A=4√(26)`

`A\approx 20`

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