geometry-and-trigonometry

[crude kraken]

so anyways

that means that the third side is 10

you know how I got that?

[prisma otter]

Nope. I just got into geo soo this is really new

[crude kraken]

Mk

[jovial phoenix]

We got a 9th grader in here boys

[prisma otter]

yep ; )

[crude kraken]

hoo

okay so

one sec lemme write it down :i

=tex B=\sin^{-1}(\frac{\sin(90)\cdot24}{26}) = 67.38

[crude kraken]

(Yea, am only in 11th :<)

[thorn talon]

Woah

[crude kraken]

\tex C = 180-A-B = 22.62

[prisma otter]

so why is 90 multiplied by 24?

[crude kraken]

Because you have to multiply sin(A)*b to get B

=tex B= \sin^{-1}(\frac{\sin(A)\cdot b}{a})

[crude kraken]

But, given this, you can use the same rule to figure out that side ab = 10, which means midpoint c is half way between that

To figure out the length of CD, you're given three important pieces of information, ac = 5, ad = 24, A = 90

using that and the law of cosines, you can figure out CD, which is 24.52

[prisma otter]

writing this down...

[crude kraken]

But if you're new to geometry idk what you're doing with a problem like this... there must be a simpler way to do this I'm sure

but idk what that is 🤷

[thorn talon]

@prisma otter what are you doing?

[prisma otter]

I've only learned quadratic formula and p therom

[crude kraken]

Find length of CD

quadratic formula doesn't go here

[prisma otter]

I am saying math class in general

[thorn talon]

Quadratic formula can be in triangles

[crude kraken]

p theorem can be used to find AB and thus CD-

[thorn talon]

But not in this case

[crude kraken]

ohhh

[prisma otter]

not in this case what?

[crude kraken]

pythagorean theorem, AB = sqrt(26^2-24^2) = sqrt(100) = 10

AB = 10

[thorn talon]

You will receive triangle questions that turn into quadratics

[crude kraken]

that means AC = 5

[thorn talon]

But not in this particular case

[crude kraken]

pythagorean theorem using AC and AD

you get your final answer

[prisma otter]

can you please walk me through i am lost like crazy

[crude kraken]

Yea

[prisma otter]

i just learned this last class thanks so much 😃

[crude kraken]

So you know that c = 26 and a = 24 right?

[prisma otter]

yes

and i got 5 for ac i understand that

[crude kraken]

Ok

So then you just use pythagorean theorem to figure out cd

[prisma otter]

and so i use 5 and 24?

[crude kraken]

=tex cd = \sqrt{24^2+5^2}

[prisma otter]

ahhhhhhhh

[crude kraken]

but if you want to sound smart and use law of sine and cosine, I can dm you what to write and you'll look smart, garunteed

[river garnet]

i like how you said that

lol

[crude kraken]

^ 👍

[prisma otter]

ok i got a decimal

of 24.51?

[crude kraken]

Mhm

And if you look at what I got earlier using sines and cosines, you'll see the answer is 24.52, almost same thing

so gg

[prisma otter]

can i have the sine and cosine too plz? XD

[crude kraken]

ye

So, law of sines (Write this down)

=tex \frac{\sin(A)}{a} = \frac{\sin(B)}{a} = \frac{\sin(C)}{a}

[prisma otter]

i have that written

[crude kraken]

k

[prisma otter]

and the b= sin^-1

that formula also

[crude kraken]

^

and also law of cosines

[thorn talon]

^

[prisma otter]

all three?

[thorn talon]

No

[crude kraken]

sin^-1 we'll get to later

[thorn talon]

It should make intiluitive sense how to use it

[prisma otter]

but which cosine?

[thorn talon]

And how to rearrange

Included angle

[crude kraken]

Just write the first line of cosine law

[thorn talon]

Angle between the two sides

[prisma otter]

cos A?

[crude kraken]

yea

coa A to cos A

[prisma otter]

and abc are all angels?

not the lengths right

[thorn talon]

Sides

lower case are sides

[crude kraken]

capital letters are angles, lowercase are sides

[prisma otter]

ohhhh

gotcha

[crude kraken]

^ probably good idea to draw that triangle

[prisma otter]

thx

[thorn talon]

Don't you get reference sheets?

[crude kraken]

I didn't

I just gotta find 'em

but if I find him one to print then I'll send it

[thorn talon]

Nice

Ur cool

[prisma otter]

its fine

[thorn talon]

I'm just too lazy to do this anymore

We all get reference sheets for basic formulas

Then we can derive from there if needed

[prisma otter]

ok i wrote the cos A

[crude kraken]

okay

now that you have those

I have to think of how to explain this to a beginner

[prisma otter]

is it just plugging in? for cos A

[crude kraken]

I assume you know algebra

[thorn talon]

What's your topic ATM?

[prisma otter]

I do Exeter so it is multiple topics

[crude kraken]

https://cdn.discordapp.com/attachments/326138757474680852/357340675517186049/Capture.png

teaching him to use law of sines and cosines

[prisma otter]

i can send you a screenshot of the page im on

[crude kraken]

pythagorean theorem is nice but hard to use here

:p

[prisma otter]

screw it i want to look smart

[crude kraken]

^

[thorn talon]

^

[crude kraken]

you probably can sleep through geometry if you get this down

[prisma otter]

ok so

[crude kraken]

like for the year

(except circles but that's different)

[thorn talon]

Geometry is eww

[crude kraken]

^

[thorn talon]

High school geometry that is

[crude kraken]

Geometry is only fun when you go 3d

[prisma otter]

thank god i found this discord

[thorn talon]

Same

I'm dying in math

[prisma otter]

so b=26 and a=24

but what is c?

wait

[crude kraken]

c is what we figure out

[prisma otter]

10 right

[crude kraken]

so you have your angle A = 90

because we have an opposite pair, we can use law of sines

[prisma otter]

we used that earlier right?

[crude kraken]

=tex \frac{\sin(90)}{24} = \frac{\sin(x)}{26}

[crude kraken]

You know how to solve this, yes?

[prisma otter]

calculator

[thorn talon]

No

[crude kraken]

no

[prisma otter]

oh crap

[thorn talon]

Cross multiply

[crude kraken]

^

[thorn talon]

Then you can calculator

[prisma otter]

ive never used sine tbh

[crude kraken]

You take the 26 and put it next to the sin(90)

[prisma otter]

k

[crude kraken]

=tex \frac{26\sin(90)}{24} = \sin(x)

[thorn talon]

You've done cross multiplication right?

[prisma otter]

yes

[crude kraken]

You can also just multiply both sides by 26..

but, now is the hard part, how do you isolate x when it's in a sin?

[prisma otter]

2390 is 90 x 26

[crude kraken]

..

[prisma otter]

wait

[crude kraken]

sin(90) is not 90

[prisma otter]

oml

[crude kraken]

The answer is inverse sin

[royal grove]

I need some help with a problem or two in my geometry test study guide

[prisma otter]

help him out^

[crude kraken]

=tex \sin^{-1}(\frac{26\sin(90)}{24}) = x

[prisma otter]

thanks for the help guys gtg

[crude kraken]

and, ask away @royal grove

See ya

I'll send you stuff :p

[royal grove]

Find the value of x for which l is parallel to m.

Lemme get a picture of the diagram

Hold on

They are squared, that’s what the numbers outside of the parenthesis are.

My teacher decided to make them super tiny

[crude kraken]

so it's

(4x-6)^2 and (3x+4)^2

?

[royal grove]

Yes

[crude kraken]

So since those are angles, you just gotta make the angles equal eachother

right?

=tex (4x-6)^2 = (3x+4)^2

[royal grove]

wait hold up

[crude kraken]

No?

[royal grove]

I sent you the wrong pic!

Sorry

[crude kraken]

gg af

[royal grove]

That’s the right one

Lmao

I just can’t figure it out

I would think they would be congruent

[crude kraken]

I think

that you just need to do

90-28

correct me if I'm wrong @people good at geometry

wait wait I'm wrong

[royal grove]

I think I have something too

[crude kraken]

180-28 = 152

[royal grove]

Wouldn’t both angles need to add up to 56?

[crude kraken]

so that angle next to 28 is 152

and I don't think that's relevant

mg lemme think again 😮

Huh.

I dunno

:e

[royal grove]

Since they need to have the same angle to be parallel?

[crude kraken]

No

to be parallel they need to have the angles between them be the same

so in a perfect world they would both be 28

but

they're not

and that's where the 56 comes in

trying to remember what to do here

OH

[river garnet]

i have a question what are the square roots of i anyways

[crude kraken]

That's for another time

I'll tell ye later

[river garnet]

lol

it is technically geometry related

[crude kraken]

okay so @royal grove what you gotta do is imagine a line going between them perpindicularly

so you solve the top triangle's angles and you know that the angle on that side of 56 is 180-90-28

so 62

x is part of the other triangle

so you solve for that

180-62-56

is 62..

[royal grove]

Okay

[crude kraken]

so now you just solve for that triangle

x = 180-62-90

which is 28

x = 28

[royal grove]

Ah

That would make since

Because of this little diagram I made

[crude kraken]

You see how I got that right?

Turning the cut into two triangles and solving the angles using right angles and such?

[royal grove]

My lines suck so sorry. xD

To create that angle the bottom line would have to move

Making it not parallel anymore

[rough juniper]

I'm not a pro, but I think lines m and l is something to use to get 180°

(for me I would find that useful

[crude kraken]

Eh

[royal grove]

Oh

That makes things easier

[crude kraken]

If you can explain it I'll swap out my triangle solution for that

[royal grove]

Since 28 degrees is supplementary to the other side

Let’s call it d

Drawing a pic

[crude kraken]

plot twist: now we're taking a derivative

[rough juniper]

hehehe

[royal grove]

28 is supp to d

[crude kraken]

Mk

[royal grove]

So d would have to be 152

[crude kraken]

Mhm

[royal grove]

Now

[rough juniper]

I think you can just do 180°-56°, since 26 obviously isn't related to 56°

[royal grove]

For it to be parallel

You need to do it to the other side??

[crude kraken]

...

[royal grove]

So it would be the same on the other side

[crude kraken]

No

because of the 52

*56

[royal grove]

Hmm

[crude kraken]

152-90 = 62, you can make another triangle that's 56, 62, 62

[rough juniper]

^

[crude kraken]

and using the same logic, you can deduce that y = 62

90-y = x

[royal grove]

I’m a bonehead so I don’t understand

[crude kraken]

90-62 = x = 28

[royal grove]

Sorry ☹️

[crude kraken]

:<

okay so

[rough juniper]

@crude kraken isn't x supplementary to 56 tho?

[crude kraken]

No

[royal grove]

Maybe I should ask about it before class

Because I have it in my brain

But for some reason

[crude kraken]

nono it's ez

[royal grove]

I can’t put it into words

That’s why i have trouble with english I

[crude kraken]

Draw this line with the right angles

and you get two triangles

the top triangle, you have two angles and can find the third

the third and 56 you can use to find the second angle on the bottom triangle

and calculate x using the right angle, the second on the bottom, and x

[royal grove]

Because the bottom angle of the top triangle, the 56 and the top angle of the bottom triangle all have to equal 180?

(Bottom triangle) So that would make 62 degrees on the top

The bottom angle would be 90

So since all triangles have to equal 180 degrees all around

90 plus 62 would be 152

Then 180 minus 152 equals 28 degrees?

Did I word it right?

[lusty sparrow]

Can somebody help me understand how to find the distance between two points?

[topaz valley]

Sure

What don't you understand?

[dark sparrow]

@lusty sparrow as in, the distance between two points using their coordinates?

or what

bc if you mean that, it's just the pythagorean theorem

[lusty sparrow]

But the distance formula is different?

[dark sparrow]

if you denote with Δx and Δy the differences in the x and y coordinates of the two points between which you want to know the distance

then you can rewrite the distance formula as

=tex D = \sqrt{\Delta x^2 + \Delta y^2}

[dark sparrow]

or alternatively

=tex D^2 = \Delta x^2 + \Delta y^2

[lusty sparrow]

Oh

[dark sparrow]

so the difference, if any, is just in how you name things

[sacred vessel]

<@&286206848099549185>

Need a and b

Area and perimeter

[dark sparrow]

broken neck

[brave drift]

You want area and perimeter of a and b right?

[upper karma]

I can't read anything, any chance you could get a better picture

[brave drift]

Area of right angled triangle is (1/2)*(base)*height

[sacred vessel]

Yee one sev

Sec

[dark sparrow]

@brave drift for a right angled triangle specifically, it's equal to half the product of its legs

[brave drift]

That is height and base, right?

[dark sparrow]

i mean you could view the legs as such

lol

[brave drift]

lol

[umbral rivet]

👀

[sacred vessel]

<@&286206848099549185>

[umbral rivet]

👀

[sacred vessel]

Srr it was sideways

I need area and peramiter

[umbral rivet]

If you want perimeter

[upper karma]

ah okay, so for A and B, do you know the formula for area of a right triangle?

[umbral rivet]

Find all the sides

[brave drift]

add the sides

[umbral rivet]

Area = multiply two sides

[brave drift]

@umbral rivet where is 1/2?

[upper karma]

perimeter is the length of the border of a shape, the length of all its side

area is what's inside the shape

so to speak

[sacred vessel]

Can you just solve it for me :p

[brave drift]

LOL

[sacred vessel]

This isnt homework i promise this is for something else

😀

Just like all the equations and what not

Please

[brave drift]

I guess no one will 'just solve '

[sacred vessel]

Then help me solve

Please

😀

This cute girl asked me to help her with maths and i dont understand this either

[brave drift]

Aah

I will solve it for you

[sacred vessel]

Thank you

[brave drift]

Perimeter of (a) -> 3sqrt(2) + 5sqrt(2) + 4sqrt(2)

That is the sum of all sides

[sacred vessel]

Yeah

And i got area

Its 12

I think

😅

[brave drift]

When you walk on the boundary of a shape, you're walking on its perimeter.

Yes.

Solve the b part with Pythagoras

[sacred vessel]

Yeah

I did that

I remembered the formulas

And it was easy at that point

Thanks though @brave drift

[brave drift]

No problem, good luck with the girl!

[sacred vessel]

Thanks 😃 😃 😃

[upper karma]

Just started geometry, forgot everything 😰

[thorn talon]

rip

[umbral rivet]

:c

[modest locust]

So I reached the point in khan academy titled "Proof with transformations" but i have no idea what these symbols are or what they mean

Think I found my answer

θ = T • H • E • T • A
Γ = G • A • M • M • A
Φ = P • H • I
ψ = P • S • I
π = P • I
τ = T • A • U
γ = G • A • M • M • A

[thorn talon]

they're just variables for the angles

[modest locust]

I see that now haha thanks

[amber vault]

as

I can speak

▱JKLM
with vertices J(2,2), K(5,3)K, L(5,−3) and M(2,−4). Round your answer to the nearest tenth

I need help

[upper karma]

So is angle 1 and 3 linear vertical or neither

[amber vault]

▱JKLM Find perimeter
with vertices J(2,2), K(5,3)K, L(5,−3) and M(2,−4). Round your answer to the nearest tenth
I need help

Pls

[thorn talon]

what issues you having?

[amber vault]

I am just

getting the answer wrong

I dont know

[thorn talon]

what have you tried?

[amber vault]

Like I know the distance formula

[thorn talon]

are you using the distance formula with the right points?

[amber vault]

Yeah

I just an area one

But like

perimeter if screwing me

did an area one*

It was easy you can just count

but perimeter is a parallelogram

I botched that up but like u get the point

[thorn talon]

well

what'd you get?

[amber vault]

Well

Um I tried a few things

12.2 18.5

Im pretty sure

[thorn talon]

2sqrt10 + 2(6)

is the answer something like that?

18.3

[amber vault]

let me try

Oh

I maybe just added wrong

Thanks

^.^

So did you just do the distance formula

and came out with that answer

[thorn talon]

yes

well not the straight up and down lines

[amber vault]

Yeah

you can just add those

What number did you come out with

3.2

Thats waht I came out with

[thorn talon]

sqrt10

don't round til the end

[amber vault]

Ok

I came out with the right answer

I wonder where I went wrong

Leave it as 3.162?

[thorn talon]

leave it as sqrt10

[amber vault]

Oh

[thorn talon]

until the end

it's exact

[amber vault]

I vaguely remember hearing about this

[thorn talon]

it's irrational

[amber vault]

hm

Didnt think about that having relevance

[thorn talon]

you can write as many decimals as you want

but it won't be exact

[amber vault]

Yeah

[amber vault]

um

Retard question

https://gyazo.com/70081e3c393aabc7ea0108eff8df653a

Why is this wrong

[rough juniper]

is the height of a triangle determined by the highest point on a triangle?

[upper karma]

y= -1x+2?

[amber vault]

^ ty

How did u get 2?

[upper karma]

If you look at the line y = x, you get a slope of 1/1

[amber vault]

I know that

[upper karma]

I flipped it by a negative

and then raised it by 2 because I needed it go through the point, -1,3

[amber vault]

I never learned that

What the fuck lol

[upper karma]

WELL that's cuz you play Dairus, play someone elegant like Orianna

😛

[amber vault]

I am kayn

I play kayn *

Currently

[upper karma]

Oh that's a little better 😛

[amber vault]

80% Wr 40 games

[upper karma]

wow is kayn that good

[amber vault]

Rhaast is so good

CC and incredible damage

Shadow assassin is never used

No tank items

other than spirit

Pls

Someone

[golden island]

should be 15.4 and 30.8

[upper karma]

Hey

[dark sparrow]

hi

can you please change your nickname to something that's easier to type?

[upper karma]

No problem

Anyway

I came here because I needed a bit of help.

[dark sparrow]

sure

what do you need help with?

[upper karma]

I need a function to calculate the area of overlap between two circles of known radii and positions.

[dark sparrow]

hmm

[upper karma]

Its for a quantum physics problem im just casually trying my hand at.

[quasi lion]

Integration?

[dark sparrow]

...care to show the problem itself?

[quasi lion]

Does the polar form ring a bell?

[upper karma]

https://www.youtube.com/watch?v=zcqZHYo7ONs

Thats the basis behind it.

[quasi lion]

Saw that vid already

[upper karma]

I'm wondering about different positionings between circles in a venn/euler diagram

and how close that can come to the experimental results.

[quasi lion]

http://mathworld.wolfram.com/Circle-CircleIntersection.html

[upper karma]

Thanks

gonna put the functions I need from there into desmos.

[quasi lion]

Shall the required knowledges be provided to the pleb

[upper karma]

I'll work it out from here, don't worry.

[river garnet]

i hate euclidean geometry so much

lol

[crude kraken]

Same :/

[river garnet]

let me try to solve it

one second

[modest locust]

quick question how do I know when to add or subtract in questions kind of threw me off

[river garnet]

@crude kraken i'm not sure

but the answer might be 48

then again i am very bad at euclidean geometry

[iron apex]

Hello

[prisma otter]

Refer to the diagram at right and find the value of x for
which triangle ABC has a right angle at C.

any1?

[umbral rivet]

ah

I see

Let me draw a picture @prisma otter

[iron apex]

Hello angel

I got need help

[umbral rivet]

👀

[umbral rivet]

what do you need help with?

aghh

;3;

I saw your question u.u

I don't think I can do that

I'll try though

[iron apex]

Alright

🤔

[umbral rivet]

let me help delta real quick

[iron apex]

ok

[umbral rivet]

his is easy to explain

[iron apex]

alright go ahead

[umbral rivet]

okie

@prisma otter

[prisma otter]

👌

thanks bud

[umbral rivet]

First, we need to solve for the missing side

And now, since we know what every side of the triangle is equal to, we can find X

[prisma otter]

ah

[umbral rivet]

Just plug everything into the pythagorean theorm

:3

This should work

I think

🤔

[prisma otter]

how did you get 36?

[umbral rivet]

6^2 = 36

[prisma otter]

ah 6 my bad

[umbral rivet]

^3^

[river garnet]

@prisma otter is your name a reference to delta balls? probably not right?

just curious

[prisma otter]

nah haha

[river garnet]

still

you should change it to that

it'd be fun

[prisma otter]

kk

[umbral rivet]

x3

If you have any more questions, feel free to ask

now @iron apex

[iron apex]

Ok

[umbral rivet]

Now, I'm pretty sure you can use the distance formula for this

but I'm just doing it the long way because uh

I forgot what the distance formula is

so basically, I'm just trying to find the length of ST and SR

in this picture, you can see that for ST, I got sqrt193

🤔

[river garnet]

pretty sure 193 is prime

i'm not certain

[umbral rivet]

👀

[river garnet]

that definitely is not any of the choices though

i mean i know 197 is

[prisma otter]

@umbral rivet how didja get 6 tho im stuck on that part

[umbral rivet]

What do you mean? :o

[river garnet]

i mean your answer can't be true lol

[umbral rivet]

aghhh

I'm messing everything up

@prisma otter I actidentally transformed my b^2 into a 6^2

[prisma otter]

ahhh

[umbral rivet]

because they look really similar

[prisma otter]

its ok gotcha

[umbral rivet]

ughhh

[prisma otter]

i can solve it its fine

thx bud

[umbral rivet]

i'm like, really awful at answering questions

<3

[prisma otter]

its fine bro dont worry my handwriting is horrible

[umbral rivet]

💕

[river garnet]

@iron apex oh wait i'm stupid you found the base

[iron apex]

wait sorry I was eating

the base?

yeah what about it?

[river garnet]

the uh previous question

[iron apex]

oh yes yes yes

[iron apex]

my fault

wait let me see if it right

[umbral rivet]

:3

[iron apex]

the answer is 45

[umbral rivet]

Now since I know ST and SR

oh

[iron apex]

I check it up the answer

[umbral rivet]

O.o

[river garnet]

oh wait i wasn't talking to you

this all just got confusing lol

[umbral rivet]

yeah

[iron apex]

Yeah that why it's common core

[umbral rivet]

wat

[iron apex]

"Common core"

[umbral rivet]

I don't know who's talking to who

[iron apex]

Geometry

[river garnet]

just refuse to do the question

that's the best answer

lol

[umbral rivet]

@iron apex

so um

This is where I'm at now

[iron apex]

no the other one is right

[umbral rivet]

?

[iron apex]

the one you did the previous work

[umbral rivet]

what do you mean

[iron apex]

this one is right

[umbral rivet]

but

what

That's not the area of the triangle

that's the side length of SR

[iron apex]

no it say

how many square units

[umbral rivet]

😢

[iron apex]

no you were right the work.

[umbral rivet]

uwu

[river garnet]

if they want the area though

isn't it sqrt(193)3sqrt(5) / 2

well

we don't know if it's a right triangle

nvm

[umbral rivet]

:3

[iron apex]

the answer is 45

you don't have to square root

[umbral rivet]

uwu

[iron apex]

Bruh

you want my geometry packet right now

I got you

[umbral rivet]

X3

If you have any more questions, I'd be glad to help :3

<3

I love math

[iron apex]

http://www.jmap.org/JMAPRegentsExamArchives/GEOMETRYCCSSEXAMS/ExamAnswers/0616ExamGEOans.pdf

go to question 22

[umbral rivet]

👀

[iron apex]

Yeah

[river garnet]

this question is dumb

[umbral rivet]

We just did that one though :3

[iron apex]

IKR

thank you spook

[river garnet]

because it assumes it's a right angle triangle

[iron apex]

Exactly!

[river garnet]

without even mentioning it is

[iron apex]

"Common Core"

"Geometry"

[umbral rivet]

uwu

[iron apex]

That why it's Common Core.

I really hate common core geometry

This is why I fail 2 time my exam

[river garnet]

not only that but as i said the base can't be sqrt(193) as 193 is prime

[iron apex]

I fail that exam two time

I fail geometry 2 time

first one was a 60

second time was a 51

[river garnet]

and when multiplied by the other side it won't give any of the answers below

[iron apex]

I was suppose to take trig this year.

Yeah

that I'm trying to say

[umbral rivet]

:(

[iron apex]

I'm confuse this question that why

wait let me check if it error

Nah it right

[river garnet]

i'm sure there is a way to solve for the area

in another way

but you know

you would need a ruler

[iron apex]

Oh

[river garnet]

lol

[iron apex]

I fail 2 time geometry

[river garnet]

even then

[umbral rivet]

🤔 I'm confused

[iron apex]

I need to pass geometry this year

or else I'm fuck

[umbral rivet]

You told me to look at number 22 :P

[iron apex]

yes

question 22

[umbral rivet]

did we not just do that one though? :p

[river garnet]

yeah no he was trying to show you the solution

[iron apex]

Yeah.

I'm really confuse the solution

[river garnet]

how they solved for it

[iron apex]

yeah

[river garnet]

so you don't understand how they solved for it?

[iron apex]

no I understand

[river garnet]

or?

ok

[iron apex]

I just want to make sure what next if you square root

[umbral rivet]

I mean

[river garnet]

then we're on the same page that this question is illogical

[iron apex]

the next step I mean

Yeah

Most of the question is illogical

[umbral rivet]

all you need to do is find all 3 sides of that triangle

multiply it by two

then divide it by two

[river garnet]

@iron apex what is your name is not illogical

@umbral rivet yes

but that's assuming the triangle is a right angle triangle

[umbral rivet]

I mean

[river garnet]

it doesn't say that in the question

[umbral rivet]

In geometry

[iron apex]

Dude let me compare the common core

to standard geometry question

[umbral rivet]

My math teacher told us to just assume it's a right triangle if it looks like one

[iron apex]

alright

[umbral rivet]

And unless it's specifically stated so, it's a right triangle

[river garnet]

@umbral rivet that's your math teacher not math itself lol

never assume facts

well never assume facts that could be true

but aren't stated

[umbral rivet]

u3u

shrug

[river garnet]

i don't atleast as that is illogical

[iron apex]

this is standard geometry

[river garnet]

it's like assuming a function is linear just because in 100 different instances it looks linear

[iron apex]

compare to common core

[umbral rivet]

👀

I did problems like that last year in geo

[iron apex]

this is common core geometry

compare to standard

oblivious cc is harder than standard

[river garnet]

and it gives valid information it looks like

[iron apex]

Yeah

I gtg

[river garnet]

there are perpendicular symbols

[iron apex]

anyway thanks for the help

[river garnet]

that other question isn't so bad either though

considering you have a rectangle

[umbral rivet]

^

And the circle and the triangle

:3

[modest locust]

I dont get this questiona t all there is nothing for me to work with or did I miss something?

[golden island]

Whats the equation with x in it?

[modest locust]

it shows no equations just a and b thats why im confused

[umbral rivet]

👀

huh

that's really weird

is this for homework?

If it is, you should take a screenshot and just send it to your teacher

:3

[modest locust]

it was on khan academy lol

[umbral rivet]

👀

[modest locust]

I might of been able to solve if there was equations lol

[umbral rivet]

xP

[modest locust]

anyways thanks guys I thought I was loosing my mind lol

[umbral rivet]

xP

<3

[wind gale]

someone help me Im getting pissed off

Arrange the natural numbers from 1 to 7 in the circles so that the sum of the numbers along each line equals 10

[timid gorge]

Anyone help me with this question

I have an answer, just not sure if it's right

[topaz valley]

What have you tried?

Or did you want us to confirm your answer

[crude kraken]

"Is the area of the two blue areas equal? Show how." A side of either square is length x.

[dark sparrow]

@crude kraken what's the area of a quarter circle with radius x?

and what's the area of a circle with diameter x?

[crude kraken]

pix^2

and, pi(x/2)^2

[dark sparrow]

pix^2
i said quarter circle

[crude kraken]

or 4pi(x^2)

pi(x^2)/4

[dark sparrow]

yeah

they're both πx^2/4

[crude kraken]

How is 4pi(x^2) = pi(x^2)/4

wait wait

oh

Got it, thanks

What I had so far :p

[dark sparrow]

so i guess you're asking why the area element in polar coordinates is r dr dθ, in contrast to the dx dy you're used to from cartesian

this isn't a very formal proof

but i hope it gets the point across

dS is supposed to stand for the infinitesimal area element, dr for the infinitesimal radial coordinate change and dθ for the infinitesimal angle coordinate change

ofc

@dark geyser does that explain it?

[crude kraken]

I thought I knew geo

until this moment

[dark sparrow]

you know how to find the area of a circular sector, right? :p

[crude kraken]

uh

area of circle

• the fraction for sector of circle

but then why not just express it like (1/n)pi(r^2)

what is d(theta)?

what's d at all?

[dark sparrow]

a calculus thing

[crude kraken]

Yea, derivatives

[dark sparrow]

differential

[crude kraken]

One day I'll understand 'em but right now I'm trying to learn limits 😛

[dark sparrow]

anyway

if you have a sector of a circle of radius r that has an angle of α, its area is αr^2/2

if α is measured in radians

[crude kraken]

Mhm?

[dark sparrow]

anyway i don't think trying to explain infinitesimals, a not really rigorous idea, would make sense right now

[crude kraken]

As x approaches 0 you mean

[dark sparrow]

i mean... a differential (like dx) sort of represents that change-approaching-0 in a variable

don't worry about it right now

[crude kraken]

mk

[timid gorge]

@topaz valley sorry I fell asleep haha I have got an answer just wanted to see if it was correct

I got 21.3cm

[dark sparrow]

== sqrt((6*pi)^2 + 10^2)

[charred spear]

21.33789489

[dark sparrow]

yup

[dim latch]

do you guys know of a good book as an introduction to the notion of curvature in space ?

[dim latch]

I've done a bit of differential geometry, like calculating the length of a parametrized curve, but it was over 1 year ago, and my memory may fail me

the notion of curvature in a parametrized 2d curve was defined, but I'd like to come to a broader and formal definition

[dim latch]

thanks.

[quick mural]

Those are all good. General terms aside from Reimann could be elliptical, hyperbolic, metric tensor, manifold, minkowski space and more. I'm sure people will add. There's also discrete space which is really weird.

@dim latch

[upper karma]

@dim latch If you are dead on set on the idea of taking a derivative in space, then I can't help, if not, then I'd recommend DIfferntial Topology by Milson it's a TERSE covering of diff. top. but yea.

[dim latch]

terse ?

[upper karma]

sparing in the use of words; abrupt.

There's little exposition besides definitions, proofs are kind of "you should be smart enough to do this yourself" flavour

[quick mural]

It seemed from the way he talked about parametrizing curves that he was around Calc III

Better safe than sorry, I suppose.

[mossy oriole]

hi

I just have one question

If I have a semi-circle with four other semi-circles inside of that semi circle (with their bottoms being portions of the bottom edge of the big semi-circle), is the sum of the arc length = the arc length of the big semicircle?

[quick mural]

@mossy oriole just the "exterior" of them or the whole perimeter of all four combined?

[slim gorge]

hey guys

i got this problem:
Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a>b)
Find the length of the segment that the diagonals intercept on the midline of the trapezoid.

i tried it, for like 20-25 mins, and then looked at the solution

but i didnt get it

the solution was :

Let P and Q be the midpoint of AB and CD; Let K and L be the intersection points of PQ with the diagonalsAC and BD, respectively. Then PL = a/2 and PK = b/2 and so KL = PL - PK = 1/2(a-b).

How did they get PL = a/2 and PK = b/2 ?

@dark sparrow

[dark sparrow]

do you have a picture?

[blissful hill]

@slim gorge mid lane of triangle equals half of base

Midline of trapezoid divides all lines connecting both bases into halves

[surreal bolt]

nice.

[blissful hill]

If PQ is midline then both diagonals are divided into halves

[slim gorge]

OH

[slim gorge]

hey

anotha problem

Points A_1 and B_1 divide sides BC and AC of triangle ABC in the ratios BA_1 : A_1 C = 1 : p and AB_1 : B_1 C = 1 : q , respectively. In what ratio is AA_1 divided by BB_1 ?

Solution:
***Denote the intersection point of AA_1 with BB_1 by O. In triangle B_1 B C draw segment A_1 A_2 so that
A_1 A_2 || B B_1. Then (B_1 C) / (B_1 A_2) = 1 + p and so

AO : O A_1
= AB_1 : B_1 A_2
= B_1 C : q * (B_1 A_2)
= (1+p) : q ***

[dark sparrow]

dude

draw a picture

it's geometry

[slim gorge]

i did

i still couldnt understand

[uncut spoke]

pls do

i have 8 pages to do

help

[fallow sapphire]

So, I was looking back into my notes and noticed that I was struggling in the section involving graphing slopes, can someone explain that to me please?

[crude kraken]

What don't you understand?

[fallow sapphire]

how to make them graphable

[crude kraken]

You mean y=mx+b right?

[fallow sapphire]

my teacher wasn't really good at telling me these things

Yeah
but I dont get what the y intercept is

[crude kraken]

The y intercept (b) is what the equation is while x equals 0

m(0) = 0, so the y intercept in mx+b is b.

[fallow sapphire]

isn't b the intercept

[crude kraken]

Yes

b is the y intercept

[fallow sapphire]

but if it = 0 doesn't it not exist?

[crude kraken]

which means the first point you plot, is at x=0, you plot it where y = b

And, it can still exist at b=0

[fallow sapphire]

ohh

like the middle?

[crude kraken]

if you have mx+0, you can still plot (0,0)

yea

[fallow sapphire]

0 = the middle okay

[crude kraken]

mhm

[fallow sapphire]

Okay this is helping

What does the M mean

[crude kraken]

Slope (m) is rise over run.

=tex m=\frac{rise}{run}

[fallow sapphire]

ohh

[crude kraken]

Rise is how high the point goes, and run is how far.

[fallow sapphire]

so if I have (6,3)

it would be this

[crude kraken]

If m is 2, you have 2/1, so you go up 2 and to the right 1

if b = 0, then yes.

6/3 = 2

[fallow sapphire]

=tex m=\frac{6(rise)}{3(run)}

[crude kraken]

I mean, 3/6 = .5

x goes first, and x is run, so it would me m = 3/6

[lost hare]

wow cameron this is awesome

whoops

[fallow sapphire]

ooh

[lost hare]

wrong server

[crude kraken]

gg

[lost hare]

lol

[fallow sapphire]

is it like this

[crude kraken]

=tex m=\frac{3}{6}

[fallow sapphire]

=tex {y}{x}

[fallow sapphire]

or the other way

[crude kraken]

It's (x,y)

So if you try to graph y=3x+2

[fallow sapphire]

=tex \text (x,y)

[crude kraken]

the first point you graph is (0,2), you know why?

[fallow sapphire]

Why?

[crude kraken]

Because when x = 0, y = 2

y = 3x + 2

3(0) + 2 = 0 + 2 = 2

[fallow sapphire]

you lost me

[crude kraken]

When you plug in 0 for x, that's how you get the y intercept

[fallow sapphire]

okay

but where did the 2 come from

oh wait

the 6/3

right?

[crude kraken]

No, I'm demonstrating if you have to graph y=3x+2

[fallow sapphire]

Oh okay

[crude kraken]

What are you trying to graph now? Probably better to demonstrate with that

[fallow sapphire]

I wasn't specifically graphing anything just asking how to do it

You can give me some numbers though

[crude kraken]

Okay

=tex y=\frac{1}{2}x+3

[crude kraken]

What's the y intercept?

[fallow sapphire]

3

[crude kraken]

So, the first point you can graph is (0,3)

[fallow sapphire]

ok

where is the 0 from? the x?

[crude kraken]

Yea

[fallow sapphire]

ok

[crude kraken]

You plug in 0 for x to find the y intercept, which means that when you graph (x,y), you get (0,y)

you told me that y is 3 when x is 0, so it's (0,3)

[fallow sapphire]

Ohhhhh

Could you plug in anything for x or just 0?

[crude kraken]

You can plug anything in

but keep it simple generally.

[fallow sapphire]

okay

[crude kraken]

So for y=1/2x + 3, what happens if you make x=1?

[fallow sapphire]

it is (1,3)?

[crude kraken]

No

[fallow sapphire]

oh

[crude kraken]

x has a value now, and when you plug it in, what does the equation look like?

[fallow sapphire]

y=1/2(1)+3?

[crude kraken]

Yes

so what's 1/2 times 1

plus 3

[fallow sapphire]

3.5?

[crude kraken]

Yea.

But it's ugly to graph half-points, so let's try the next one, 2.

What's y when x=2?

[fallow sapphire]

y=1/2(2)+3

so its .5*2+3?

[crude kraken]

Mhm

[fallow sapphire]

so 4

[crude kraken]

Yes, so the next point is (2,4)

So you've marked (0,3) and (2,4), and because the equation is to graph a line, you can line up a ruler to the two points and draw your line

[fallow sapphire]

I get it now!

[crude kraken]

Mhm

Now take note of how the point moved

[fallow sapphire]

.5

[crude kraken]

What's the difference between (0,3) and (2,4)?

[fallow sapphire]

(2,1)?

[crude kraken]

Mhm, so you can say the point moved to the right 2 and up 1, right?

[fallow sapphire]

Rise and run yes

[crude kraken]

Mhm

So,

[lost hare]

is everyone here teachers or just smart as hell?

[crude kraken]

=tex \frac{rise}{run} = \frac{1}{2} = \text{ Up 1, right 2}

[crude kraken]

I'm neither a teacher nor smart, I just happen to know how to graph so I'm helping out someone who doesn't know.

[fallow sapphire]

Ohhh