anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
http://learn.flvs.net/webdav/assessment_images/educator_geometry/v15/module06/0600_g12_q4.jpg
anonymous
  • anonymous
@Here_to_Help15
Here_to_Help15
  • Here_to_Help15
I um failed but @johnweldon1993 it the best user in OS hes the recommendation i was talking about lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
okay haha thank you
johnweldon1993
  • 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
  • anonymous
oh crap dude im sorry i coul have screen shotted it for you. sorry u had to draw it out
johnweldon1993
  • johnweldon1993
Let me label it with the ABCD lol |dw:1433883546238:dw|
johnweldon1993
  • johnweldon1993
Lol no worries...at least there weren't like decimals or crap like that lol
anonymous
  • anonymous
haha right!!
johnweldon1993
  • 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
  • anonymous
yes
johnweldon1993
  • johnweldon1993
Okay...so now |dw:1433883752245:dw| Look familiar?
anonymous
  • anonymous
umm kinda yea
johnweldon1993
  • 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
  • anonymous
yes that makes sence
johnweldon1993
  • 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
  • johnweldon1993
So \[\large Perimeter = \sqrt{5} + \sqrt{5} + \sqrt{10} + \sqrt{10}\] \[\large Perimeter = 2\sqrt{5} + 2\sqrt{10} = ?\]
anonymous
  • anonymous
i got 10.79 so it would be 10.8? SO A??
johnweldon1993
  • johnweldon1993
Correct :)
anonymous
  • anonymous
yess thank you so much!!!!
johnweldon1993
  • johnweldon1993
No problem :)
anonymous
  • anonymous
Can u help me with another?
johnweldon1993
  • johnweldon1993
Maybe...not very good in math :P lol
anonymous
  • anonymous
haha umm what? r u crazy? u r really good at math
johnweldon1993
  • johnweldon1993
Lol I try :P
anonymous
  • anonymous
1 Attachment
johnweldon1993
  • johnweldon1993
Okay so Arc length can be found using the formula \[\large s = r\theta\]
anonymous
  • anonymous
okay
Here_to_Help15
  • Here_to_Help15
Told ya he was good lol
anonymous
  • anonymous
haha i know @Here_to_Help15 thank you alos very very much
johnweldon1993
  • 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
  • Here_to_Help15
Dont thank me thank him ^ lol
Here_to_Help15
  • Here_to_Help15
and pfft not good at math thats a good one lol
anonymous
  • anonymous
hold on i think i did this wrong... im trying to get this into my calculator but its being dumb
anonymous
  • anonymous
i got 1.13....
johnweldon1993
  • johnweldon1993
Not quite :) Want an easier hint? \[\large \cancel{6}\times \frac{\pi}{\cancel{6}} \]
anonymous
  • anonymous
ohhh haha wow i am dumb i got it now thank you.....
johnweldon1993
  • johnweldon1993
No you're not...you absolutely can go through the calculator...but why go through all that when there are shortcuts :P
anonymous
  • 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
  • Here_to_Help15
\(\Huge\tt\color{#f9bec7}{He's~Right~ lol }\)
johnweldon1993
  • 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
  • anonymous
1 Attachment
anonymous
  • anonymous
this is the last one i need help with!!!
Here_to_Help15
  • Here_to_Help15
\(\Huge\tt\color{#3c0808}{Can~i ~Borrow~you~now~Lol !}\)
anonymous
  • anonymous
@johnweldon1993
anonymous
  • anonymous
u can help @Here_to_Help15 fist though
Here_to_Help15
  • Here_to_Help15
\(\Huge\tt\color{#f9bec7}{Ladies~ First :) }\)
johnweldon1993
  • johnweldon1993
Oh geez another one!? ;D haha just kidding Okay... |dw:1433885518952:dw|
johnweldon1993
  • johnweldon1993
With a right triangle...what do we know about sin and cos?
anonymous
  • anonymous
haha im sorry last one i promis :p
Here_to_Help15
  • Here_to_Help15
Can i join this post i am also learning this :)
johnweldon1993
  • 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
  • Here_to_Help15
Sine Function: sin(θ) = Opposite / Hypotenuse Cosine Function: cos(θ) = Adjacent / Hypotenuse Tangent Function: tan(θ) = Opposite / Adjacent
anonymous
  • anonymous
sin = opposite over hypotenuse cos= adjacent over hypotenuse
Here_to_Help15
  • Here_to_Help15
I might as well throw tangent in the mix as well lol
johnweldon1993
  • 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
  • johnweldon1993
clearly I cannot spell -_- lol
Here_to_Help15
  • Here_to_Help15
haha my teacher just literally did a livelesson on this at 12 pm
Here_to_Help15
  • Here_to_Help15
So then that means you need a calculator
johnweldon1993
  • johnweldon1993
No not at all (here anyways) this question only wants the ratios....which that is :)
anonymous
  • anonymous
wait wouldnt it be A?
Here_to_Help15
  • Here_to_Help15
Oh lol i thought it was going more further
johnweldon1993
  • 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
  • Here_to_Help15
lol
anonymous
  • 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
  • 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
  • Here_to_Help15
\(\Huge\tt\color{#82ffd6}{Oh~I~See!}\)
Here_to_Help15
  • Here_to_Help15
sorry guys i love doing latex lol
johnweldon1993
  • johnweldon1993
\[\Huge\tt\color{#82ffd6}{OIC!}\]
anonymous
  • anonymous
haha yes i did need your help :p your drawing skills uff ya know they pef hahahaha
Here_to_Help15
  • Here_to_Help15
\(\large\tt\color{#99d3f0}{LIKE~IK~RIGHT! }\)
johnweldon1993
  • johnweldon1993
XD lol mmhmm we're gonna be good friends @CayleeS23 :P
Here_to_Help15
  • Here_to_Help15
\(\Huge\tt\color{#f9bec7}{Hahaha!}\)
anonymous
  • anonymous
haha i thin soo ;)
anonymous
  • anonymous
think* damn i cant type either
johnweldon1993
  • johnweldon1993
As long as you dont.. <.< >.> Stalk me :O XD no jk that's invited too XDDD
anonymous
  • anonymous
hahahha oh my!! imma dm you
Here_to_Help15
  • Here_to_Help15
\(\Huge\tt\color{#f9bec7}{Im~so~lonely~got~nobody~*cries* }\)
anonymous
  • anonymous
whattt nooo u got me @Here_to_Help15 your chill man!
Here_to_Help15
  • Here_to_Help15
\(\Huge\tt\color{#f9bec7}{*Heart~Breaks
anonymous
  • anonymous
brb i have like 2 questions then i can turn this in and then i can chat with yall!!
johnweldon1993
  • 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
  • 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
  • Here_to_Help15
How do you that one lol @johnweldon1993
johnweldon1993
  • johnweldon1993
Secrets :P
Here_to_Help15
  • Here_to_Help15
Argh i tend to discover them lol
Here_to_Help15
  • Here_to_Help15
Out of the blue question but john do you play clash of clans ?
anonymous
  • anonymous
omg poeple on here r crazy af.. report that person
Here_to_Help15
  • Here_to_Help15
lol i told him "is that a virtual threat because im loving it :)"
anonymous
  • anonymous
hahha yess thats great
johnweldon1993
  • johnweldon1993
Lol and no I dont play clash of clans @Here_to_Help15
Here_to_Help15
  • Here_to_Help15
John you happen to speak french?
johnweldon1993
  • 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
  • Here_to_Help15
LOL

Looking for something else?

Not the answer you are looking for? Search for more explanations.