Find the perimeter of the following shape:
Shape ABCD is shown. Point A is at 7, 5. Point B is at 6, 3. Point C is at 3, 2. Point D is at 4, 4.
10.8
11
11.4
11.6

- anonymous

- jamiebookeater

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- anonymous

http://learn.flvs.net/webdav/assessment_images/educator_geometry/v15/module06/0600_g12_q4.jpg

- anonymous

- Here_to_Help15

I um failed but @johnweldon1993 it the best user in OS hes the recommendation i was talking about lol

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## More answers

- anonymous

okay haha thank you

- johnweldon1993

Lol I appreciate that @Here_to_Help15
So sadly the link wont load for me however let me draw a rough sketch
|dw:1433883395422:dw|

- anonymous

oh crap dude im sorry i coul have screen shotted it for you. sorry u had to draw it out

- johnweldon1993

Let me label it with the ABCD lol
|dw:1433883546238:dw|

- johnweldon1993

Lol no worries...at least there weren't like decimals or crap like that lol

- anonymous

haha right!!

- johnweldon1993

Okay...so we need the distance between all these points
Lets work with just points A and B for now
Point A is at (7,5)
Point B is at (6,3)
How far down and how far over from point A do you have to go to get to point B?
You have to go down 2 units...and over to the left 1 unit right?

- anonymous

yes

- johnweldon1993

Okay...so now
|dw:1433883752245:dw|
Look familiar?

- anonymous

umm kinda yea

- johnweldon1993

It's a right triangle
Which means we can use the Pythagorean Theorem to solve for that missing length
\[\large a^2 + b^2 = c^2\]
we need 'c' so
\[\large c = \sqrt{a^2 + b^2}\]
which here would be
\[\large c = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]
That is the distance between A and B...did that process make sense?

- anonymous

yes that makes sence

- johnweldon1993

Okay good...so we just do that for each of these points...
Distance from A to B
2 down, 1 over ...then just \(\large c = \sqrt{2^2 + 1^2} = \sqrt{5}\)
Distance from B to C
1 down, 3 over so \(\large c = \sqrt{1^2 + 3^3} = \sqrt{10}\)
Distance from C to D
2 up, 1 over so \(\large c = \sqrt{2^2 + 1^2} = \sqrt{5}\)
and Distance from D to A
1 up, 3 over so \(\large c = \sqrt{1^2 + 3^2} = \sqrt{10}\)
SO! Now that we have all those distances..
|dw:1433884295940:dw|
We just add them all together to get the perimeter

- johnweldon1993

So
\[\large Perimeter = \sqrt{5} + \sqrt{5} + \sqrt{10} + \sqrt{10}\]
\[\large Perimeter = 2\sqrt{5} + 2\sqrt{10} = ?\]

- anonymous

i got 10.79 so it would be 10.8? SO A??

- johnweldon1993

Correct :)

- anonymous

yess thank you so much!!!!

- johnweldon1993

No problem :)

- anonymous

Can u help me with another?

- johnweldon1993

Maybe...not very good in math :P lol

- anonymous

haha umm what? r u crazy? u r really good at math

- johnweldon1993

Lol I try :P

- anonymous

##### 1 Attachment

- johnweldon1993

Okay so Arc length can be found using the formula
\[\large s = r\theta\]

- anonymous

okay

- Here_to_Help15

Told ya he was good lol

- anonymous

haha i know @Here_to_Help15 thank you alos very very much

- johnweldon1993

BUT! 1 thing that may trip you up is we need the angle in radiens
to convert the angle to radiens we need to do \(\large \theta \times \frac{\pi}{180}\)
so here \(\large 30 \times \frac{\pi}{180} = \frac{30\pi}{180} = \frac{\pi}{6}\)
So all we need to do is
\[\large s = r\theta = 6ft \times \frac{\pi}{6} = ?\]

- Here_to_Help15

Dont thank me thank him ^ lol

- Here_to_Help15

and pfft not good at math thats a good one lol

- anonymous

hold on i think i did this wrong... im trying to get this into my calculator but its being dumb

- anonymous

i got 1.13....

- johnweldon1993

Not quite :)
Want an easier hint?
\[\large \cancel{6}\times \frac{\pi}{\cancel{6}} \]

- anonymous

ohhh haha wow i am dumb i got it now thank you.....

- johnweldon1993

No you're not...you absolutely can go through the calculator...but why go through all that when there are shortcuts :P

- anonymous

haha yes u just set it up in a way that i understand the second time!! thank you soooooo freaking much u have no idea how much of a help u are!!!!

- Here_to_Help15

\(\Huge\tt\color{#f9bec7}{He's~Right~ lol }\)

- johnweldon1993

Well that's what I'm here for :) if you need anything else feel free to message me or tag me in a question :)

- anonymous

##### 1 Attachment

- anonymous

this is the last one i need help with!!!

- Here_to_Help15

\(\Huge\tt\color{#3c0808}{Can~i ~Borrow~you~now~Lol !}\)

- anonymous

- anonymous

u can help @Here_to_Help15 fist though

- Here_to_Help15

\(\Huge\tt\color{#f9bec7}{Ladies~ First :) }\)

- johnweldon1993

Oh geez another one!? ;D haha just kidding
Okay...
|dw:1433885518952:dw|

- johnweldon1993

With a right triangle...what do we know about sin and cos?

- anonymous

haha im sorry last one i promis :p

- Here_to_Help15

Can i join this post i am also learning this :)

- johnweldon1993

Lol you already are part of the post :P but just work along and then we can work with yours :D

- Here_to_Help15

Sine Function:
sin(Î¸) = Opposite / Hypotenuse
Cosine Function:
cos(Î¸) = Adjacent / Hypotenuse
Tangent Function:
tan(Î¸) = Opposite / Adjacent

- anonymous

sin = opposite over hypotenuse
cos= adjacent over hypotenuse

- Here_to_Help15

I might as well throw tangent in the mix as well lol

- johnweldon1993

Perfect @CayleeS23
Okay...so in your question...it asks for sin(x)
So we look at your triangle...go to the 'x' degree....and look what is opposite that angle...and look at what the hypotenuse is
so
|dw:1433885706708:dw|
So that would mean
\[\large sin(x) = \frac{opposite}{hypotenuse} = \frac{3}{5}\]

- johnweldon1993

clearly I cannot spell -_- lol

- Here_to_Help15

haha my teacher just literally did a livelesson on this at 12 pm

- Here_to_Help15

So then that means you need a calculator

- johnweldon1993

No not at all (here anyways) this question only wants the ratios....which that is :)

- anonymous

wait wouldnt it be A?

- Here_to_Help15

Oh lol i thought it was going more further

- johnweldon1993

So that is the process to find the sin(x)
@CayleeS23 now you show me how to find cos(y)
*or just answer it correctly..pfft okay >.< XD Yes A is correct :)

- Here_to_Help15

lol

- anonymous

haha sorry!! i got it once u posted how to find sin i just did the same with cos using its rule then looked back at the answers :p i would draw it out like u but i have no idea how

- johnweldon1993

Yeah it takes some pretty advanced drawing skills ;) XD
But either way awesome job! See you didnt even need my help :P

- Here_to_Help15

\(\Huge\tt\color{#82ffd6}{Oh~I~See!}\)

- Here_to_Help15

sorry guys i love doing latex lol

- johnweldon1993

\[\Huge\tt\color{#82ffd6}{OIC!}\]

- anonymous

haha yes i did need your help :p your drawing skills uff ya know they pef hahahaha

- Here_to_Help15

\(\large\tt\color{#99d3f0}{LIKE~IK~RIGHT! }\)

- johnweldon1993

XD lol mmhmm we're gonna be good friends @CayleeS23 :P

- Here_to_Help15

\(\Huge\tt\color{#f9bec7}{Hahaha!}\)

- anonymous

haha i thin soo ;)

- anonymous

think* damn i cant type either

- johnweldon1993

As long as you dont..
<.< >.>
Stalk me :O XD no jk that's invited too XDDD

- anonymous

hahahha oh my!! imma dm you

- Here_to_Help15

\(\Huge\tt\color{#f9bec7}{Im~so~lonely~got~nobody~*cries* }\)

- anonymous

whattt nooo u got me @Here_to_Help15 your chill man!

- Here_to_Help15

\(\Huge\tt\color{#f9bec7}{*Heart~Breaks

- anonymous

brb i have like 2 questions then i can turn this in and then i can chat with yall!!

- johnweldon1993

\[\bbox [10pt, black, border:5pt solid blue]{\Huge\cal\color{red}{\bigstar}\color{orange}{I}\color{yellow}{T}\space \color{green}{W}\color{blue}{I}\color{indigo}{L}\color{violet}{L}\space\color {red}{B}\color{orange}{E}\space \color{yellow}{O}\color{green}{K}\color{blue}{A}\color{indigo}{Y}\space \color{violet}{H}\color{red}{A}\color{orange}{H}\color{yellow}{A}\color{red}{\bigstar}}\]

- Here_to_Help15

OMG LOOK WHAT SOMEONE SENT ME
please help me our else i will go to the open study board and say your name that you didn't help.So please help me in english please right now.

- Here_to_Help15

How do you that one lol @johnweldon1993

- johnweldon1993

Secrets :P

- Here_to_Help15

Argh i tend to discover them lol

- Here_to_Help15

Out of the blue question but john do you play clash of clans ?

- anonymous

omg poeple on here r crazy af.. report that person

- Here_to_Help15

lol i told him "is that a virtual threat because im loving it :)"

- anonymous

hahha yess thats great

- johnweldon1993

Lol and no I dont play clash of clans @Here_to_Help15

- Here_to_Help15

John you happen to speak french?

- johnweldon1993

I took 3 years of French in High school but other than introducing myself I would die quickly in France XD

- Here_to_Help15

LOL

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