Question

The rectangle ABCD is centered at the origin. C has the coordinates (w, –v). Determine the length of a diagonal of ABCD.
A.
2004-06-02-04-00_files/i0080001.jpg
B.
2004-06-02-04-00_files/i0080002.jpg
C.
2004-06-02-04-00_files/i0080003.jpg
D.
2004-06-02-04-00_files/i0080004.jpg

Answer 1

it is b go with it 
go with it

Answer 2

The length of the diagonal of rectangle ABCD is $2\sqrt{v^2 + w^2}$

The coordinate C is given as:

$C =(w,-v)$

The rectangle is said to be centered at the origin.

So, we have the adjacent coordinates of C are:

$B = (w,v)$

$D = (-w,-v)$

Calculate lengths BC and CD using the following distance (d) formula

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

So, we have:

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

$BC = \sqrt{0^2 + (2v)^2}$

$BC = \sqrt{(2v)^2}$

$BC = 2v$

$CD = \sqrt{(w --w)^2 + (-v- -v)^2}$

$CD = \sqrt{(2w)^2 + (0)^2}$

$CD = 2w$

The length (L) of the diagonal is then calculated using the following Pythagoras theorem:

$L =\sqrt{ BC^2 + CD^2}$

So, we have:

$L =\sqrt{ (2v)^2 + (2w)^2}$

$L =\sqrt{ 4v^2 + 4w^2}$

Factor out 4

$L =\sqrt{ 4(v^2 + w^2)}$

Take positive square root of 4

$L =2\sqrt{v^2 + w^2}$

Hence, the length of the diagonal of rectangle ABCD is $2\sqrt{v^2 + w^2}$

Read more about diagonals at:

https://brainly.com/question/17117320