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:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(2-0)^2+(4-1)^2}$

$AB=\sqrt{(2)^2+(3)^2}$

$AB=\sqrt{4+9}$

$AB=\sqrt{13}$

Similarly,

$BC=\sqrt{(6-2)^2+(0-4)^2}$

$BC=\sqrt{(4)^2+(-4)^2}$

$BC=\sqrt{16+16}$

$BC=\sqrt{32}$

$BC=4\sqrt{2}$

Now, the length of the rectangle is $AB=\sqrt{13}$ and the width of the rectangle is $BC=4\sqrt{2}$. So, the area of the rectangle is:

$A=length \times width$

$A=\sqrt{13}\times 4\sqrt{2}$

$A=4\sqrt{26}$

$A\approx 20$

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