Question

The coordinates of the vertices of a rectangle are (−3, 4), (7, 2) , (6, −3), and (−4, −1) .

What is the perimeter of the rectangle to the nearest tenth of a unit?



Enter your answer as a decimal in the box. Round each step to the nearest tenth.

Answer 1

Use distance formula...

\(\huge\bf{d=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}}\)

Find the distance between `(-3,4) and (7,2) then multiply by 2.` Then find the distance between `(7,2) and (6,-3) then multiply by 2.` Once this is done you would add the two distances and find your answer.

Answer 2

how do i do that?

Answer 3

Plug in the points...

\(\huge\bf{d=\sqrt{(2-4)^{2}+(7-(-3))^{2}}}\)

Answer 4

i dont know how to do that.... im not very good at math..

Answer 5

We need to simplify...so we would subtract the parenthesis...

\(\huge\bf{d=\sqrt{(-2)^{2}+(10)^{2}}}\) Next we square them...

\(\huge\bf{d=\sqrt{4+100}}\) Add..

\(\huge\bf{d=\sqrt{104}=2\sqrt{26}}\) We would multiply this by 2...

\(\huge\bf{(2\sqrt{26})\times 2 = 4\sqrt{26}}\)

This is the distance total of two opposite sides...

Answer 6

Now we find the next two points :)

\(\huge\bf{d=\sqrt{(-3 - 2)^{2}+(6-7)^{2}}}\) Subtract...

\(\huge\bf{d=\sqrt{(-5)^{2}+(1)^{2}}}\) Square...

\(\huge\bf{d= \sqrt{25 + 1}}\) Add..

\(\huge\bf{d= \sqrt{26}}\) In which we would multiply by 2....

\(\huge{(\sqrt{26}) \times 2 = 2\sqrt{26}}\)

Answer 7

You would then add the two....

\(\huge\bf{2 \sqrt{26} + 4 \sqrt{26} = ?}\)

Answer 8

@leeanm

Answer 9

Plug this in to your calculator....

Answer 10

30.5?

Answer 11

Close. If you plug into your cal you would get `30.59411708` you would need to round to the nearest tenth so it would be `30.6`

Answer 12

so 30.6 is my answer?

Answer 13

Yes

Answer 14

ok!

Answer 15

Can you help me with another?

Answer 16

I can't. Sorry I have to go though I think @Will.H can he's one great owlet :)

Answer 17

ok

Answer 18

But post it in a new post to not crowd this one....Oh and...


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Answer 19

oh amazing work @563blackghost

Answer 20

Why thank you @Will.H :)

Answer 21

well that's a really remarkable work!
great job :)

Answer 22

@leeanm if you have other question post them up
in a different post and tag me like @Will.H