Question

Can you check this question please?

Answer 1

Check please?
@Hero

Answer 2

so... hmmm how do you find the area of a rectangle?

Answer 3

I just counted the squares... @jdoe0001

Answer 4

ok...hmm so..hmm what did you get anyway?

Answer 5

Did you not look at the image I attached? 52...

Answer 6

@jdoe0001

Answer 7

hm nope, because there wasn't any

Answer 8

@jdoe0001 I attached an image, what are you talking about. Anyway did I get it right?

Answer 9

I don't see any attachment there

so. dunno

but I guess counting them is one way to do it

Answer 10

if you meant the image with the picture of the rectangle
I saw that, sure

Answer 11

but not any picture with your count

so.. what did you get anyway?

Answer 12

@Hero I got 52. Is that right?

Answer 13

ahemm nope

Answer 14

so.... how would you get the area of a rectangle anyway?

say this one

Answer 15

LxW
So I used a calculator and got 66.

Answer 16

@jdoe0001

Answer 17

well
you seem to be ... like hmm quite lagged or something

Answer 18

L x W
so length * width

so let's take a peek at your picture then


let's say to get our width
let us pick the two points of 2,3 and 5,-1

and for the length, say 2,3 and -6, -3

thus \(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 3}})\quad
% (c,d)
&({\color{red}{ 5}}\quad ,&{\color{blue}{ -1}})\impliedby width\\
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 3}})\quad
% (c,d)
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\impliedby length
\end{array}
\\ \quad \\

% distance value
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

Answer 19

once you find the width, and length
use the L x W then, and that'd be the area of the rectangle

Answer 20

hmmm should be -6
so \(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 3}})\quad
% (c,d)
&({\color{red}{ 5}}\quad ,&{\color{blue}{ -1}})\impliedby width\\
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 3}})\quad
% (c,d)
&({\color{red}{ -6}}\quad ,&{\color{blue}{ -3}})\impliedby length
\end{array}
\\ \quad \\

% distance value
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)