Algebra

[autumn anvil]

Can anyone solve this

[sand gust]

Any extra things?? Like something equal to something? And also the shape having 20 cm is a rectangle?

[pseudo umbra]

you need to use pitagoras

[autumn anvil]

How

[pseudo umbra]

one edge is x-6

the other is 6

oh but you have other given?

[autumn anvil]

I got 19.1

It's wrong

400 = 36 + x^2

x^2 = 400 - 36

x^2 = 364

[pseudo umbra]

can you take a pic of the page?.i think smth is missing

[sand gust]

That shape having 20 cm is concerning

If it's a rectangle then there is hope

[pseudo umbra]

but i dont think it will be helpful

[sand gust]

We can take the whole triangle with x

[pseudo umbra]

oh thats true. then youll have 2 unknowns

x and y

[sand gust]

Wait is the answer 12?

@autumn anvil

[autumn anvil]

Idk ans

[sand gust]

Bruh

[autumn anvil]

It's 17

[sand gust]

Using trigonometric value seems the way

[autumn anvil]

Ooh ok

[pseudo umbra]

wait how?you dont have angles

[autumn anvil]

90

Rectangle

[sand gust]

As u can see there is a square

[autumn anvil]

Thanks

[sand gust]

Btw is the shape having 20 cm is rectangle?

[autumn anvil]

No it's 20

[pseudo umbra]

but how does it help me with the triangle?. i stiil wont know the nagles

[autumn anvil]

Yes

[sand gust]

Exact figure can u provide?

[autumn anvil]

My sister came and told me to do this I will provide latet

[sand gust]

Is the answer really 17?

[pseudo umbra]

i think you can only solve this with pitagoras

with 2 unknowns

[desert tulip]

You can use similarity

[pseudo umbra]

yes, but you still youll have 2 unknowns

[wheat sun]

just use similar triangles

[autumn anvil]

How

[wheat sun]

these two are similar so the proportions of the sides must be equal with the other triangles

lets call these x and 20-x

oh well x is already used

uhhh

use h ig

[autumn anvil]

Ok

[pseudo umbra]

ohhh

[wheat sun]

becaus ethes etriangles are similar

you can find h

and you can find how long the hyp of the triangles are

then you can just find the angle and find x

with some trig stuff

[pseudo umbra]

wait why 20-x?

[wheat sun]

the whole length of the rectangle on top is 20

we can say that one part is 20-h and the other part is long h

summing together they are still 20

[pseudo umbra]

you cant call it x cause you already have x

[wheat sun]

yeah i renamed it h

[pseudo umbra]

oh ok

[autumn anvil]

Can u guys solve it

[wheat sun]

no

you do it

[autumn anvil]

I just can do it

[wheat sun]

this should be enough of a help

lemme rewrite it

oh hold on wrong side

[wheat sun]

ye i think there is too little information the rectangle on top can be inclined in any way and the height is different in each case

does the problem give anything else

[pseudo umbra]

u dont need a height to solve this

[wheat sun]

im referring to x

[wheat sun]

you can also have the rectangle like this with length 20

and the x is smaller

there is not enough info

[pseudo umbra]

whats wrong with this?. if the info wiil be the same i dont see the problem

[wheat sun]

the x is different

[pseudo umbra]

its ok. if the x=6 because of this then ur right, but it isnt

[wheat sun]

do you understand what im saying

read again

[pseudo umbra]

i read. im not sure if i have an answer

[wheat sun]

this orange line and the green line have different lengths

with the same info given

we cannot find x

[pseudo umbra]

can you picture the question from the book ?. it will be much easier

[autumn anvil]

Ok

[pseudo umbra]

youre right that the green and the orange lines arent the same, but you can still find x. i hope i understoon your question

[wheat sun]

you cant

you can find x in terms of a variable

but not an exact value

[pseudo umbra]

what do you mean by that?

[wheat sun]

how can you find the value of something that can be anything

think

[pseudo umbra]

as i wrote before, you can do pitagoras in 3 triangles

[wheat sun]

then do it

[pseudo umbra]

ok

I got that x=16.82

its really close to the answer

[wheat sun]

and how did u get to that answer

[pseudo umbra]

by that. i did that on the 2 lil triangles and then on the large one. i did it by making 2 equations

[wheat sun]

show...

[autumn anvil]

Volume

[wheat sun]

can you send the exercise

[pseudo umbra]

i dont know how to explain myself, but i hope that helps

[wheat sun]

it would be (6+y)^2

and that gives you 2 variables

so there is no solution

you can use pythagora or similar triangles

you have 2 variables

[pseudo umbra]

thats what i ment to do

[pseudo umbra]

thats why you need to built 2 equations

[wheat sun]

and what would your other equation be

[pseudo umbra]

y^2+6^2=20^2-(x-6)^2+6^2

[wheat sun]

its wrong

[pseudo umbra]

why?

[wheat sun]

explain your right hand side what you did

[pseudo umbra]

i did pitagoras on the lower triangle

[wheat sun]

it would be y^2 + 6^2 = h^2

you have another variable

[pseudo umbra]

where is h in here?

[pseudo umbra]

ohh. its the same idea. youll have to put the Expression i made insted of this

[wheat sun]

your expression doesnt make sense

thats the thing

[pseudo umbra]

why noy?

[wheat sun]

explain what you did

what squares did you consider for h

[pseudo umbra]

20^2-(x-6)^2+6^2=h^2

[wheat sun]

you realize its wrong

which sides are you considering

for h^2

im not gonna keep arguing with you

you are not explaining what you're doing

[pseudo umbra]

what i wrote is the same as you wrote. i got on the first pitagoras this: x^2+(6+y)^2.

now i need the other hypotenuse (on the lower triangle), so i need to subtract this expression from 20

[wheat sun]

i said which sides are you considering

you are saying that this huge shape is the same as this small square

do you really think they have the same area

ik about what process you are doing but you are selecting the wrong sides

[pseudo umbra]

im not talking about the squares.

[wheat sun]

dude

im not gonna talk with you further with this

you are saying that 1 = 5 basically

[pseudo umbra]

what?. sorry i just cant understand u 🥲

[wheat sun]

if you have no idea what you're doing dont confuse people with your wrong ideas

not an insult

but you are making things worse

you dont even know which sides you are considering to build the purple area

[pseudo umbra]

i know what im doing i just dont understand u, and i tried to

[wheat sun]

what are you considering

im not gonna talk

this problem has no solutions

[pseudo umbra]

ok have a good year 🙂

[autumn anvil]

Happy new year

Guys

[wise spire]

seen that on tiktok

[wanton sky]

Simplifying the diagram somewhat, and adding some lengths

with unknowns $d$ and $y$, don't we have three right-angled

triangles, and hence three equations:
\begin{align*} (x-6)^2+6^2&=(20-y)^2 \qquad \text{. . . . . (1)} \ 6^2 + d^2 &= y^2 \qquad \text{. . . . . (2)} \ x^2 + (6+d)^2 &= 20^2 \qquad \text{. . . . . (3)} \end{align*}

[rustic needle]

Jay

[wanton sky]

\begin{align*} \text{From (2):} \quad d &= \sqrt{y^2-36} \qquad \text{as $d,,y > 0$} \ \text{So, in (3):} \quad x^2 + \left(6 + \sqrt{y^2-36}\right)^2 &= 400 \ x &= \sqrt{400 - \left(6 + \sqrt{y^2-36}\right)^2} \qquad \text{as $x > 0$} \ \text{So, in (1):} \quad \left[\sqrt{400 - \left(6 + \sqrt{y^2-36}\right)^2} - 6\right]^2 + 36 &= (20 - y)^2 \ 400 - \left(6 + \sqrt{y^2-36}\right)^2 - 12\sqrt{400 - \left(6 + \sqrt{y^2-36}\right)^2} + 36 &= 400 - 40y + y^2 - 36 \ -12\sqrt{y^2-36} - y^2 - 12\sqrt{400 - \left(6 + \sqrt{y^2-36}\right)^2} &= - 40y + y^2 - 72 \ 6\sqrt{y^2-36} + 6\sqrt{400 - \left(6 + \sqrt{y^2-36}\right)^2} &= 20y - y^2 + 36 \ \ \text{...which has one solution for $x,,y,,d >0$:}\quad y &\approx 6.72642...\ \text{cm} \ x&\approx 17.8401...\ \text{cm} \ d &\approx 3.0404...\ ]text{cm} \end{align*}

[rustic needle]

Jay

[wanton sky]

This is possinly (probably?) solved less messily using trigonometry, as:
\begin{align*} \frac{x-6}{6}&=\frac{6}{d} = \frac{x}{6+d} \ 6+d &= \frac{6x}{x-6} \ \text{And,} \quad \frac{6}{20-y}&=\frac{d}{y} = \frac{6+d}{20} \ \implies 20 - y &= \frac{6\times 20}{6+d} \ &= 120\cdot \frac{x-6}{6x} \ &= \frac{20(x-6)}{x} \ \text{Making (1):} \quad (x-6)^2 + 36 &= \left(\frac{20(x-6)}{x}\right)^2 \end{align*}

[rustic needle]

Jay

[wanton sky]

And the exact solution is then $x = 3 + \sqrt{109} +\sqrt{82-6\sqrt{109}}$

[rustic needle]

Jay

[wanton sky]

And a second solution is $x = 3 + \sqrt{109} - \sqrt{82-6\sqrt{109}} \approx 9.0405...$,

in which case $y = \frac{120}{x} \approx 13.2735...$

and $d = \sqrt{y^2 - 36} \approx 11.84009...$

[rustic needle]

Jay

[wanton sky]

Note... rearranging my earlier trig statement that:
\begin{align*} 20 - y &= \frac{20(x-6)}{x} \ 20x - xy &= 20x - 120 \ xy &= 120 \end{align*}
Means that we are solving simultaneously for the intersection

of the rectangular hyperbola $xy=120$

with the hyperbola $(x - 6)^2 + 6^2 = (20 - y)^2$

with the two (valid) solutions being $\left(x - 3 - \sqrt{109}\right)^2 = 82 - 6\sqrt{109}$

and the two invalid solutions being $\left(x - 3 + \sqrt{109}\right)^2 = 82 + 6\sqrt{109}$.

The invalid solutions have $x < 6$, note, which is prohibited by the

constraints imposed by the question.

[rustic needle]

Jay

[sand gust]

Literally a elementary level question having this much hardness means there is something wrong in this question

[pseudo umbra]

It isn't hard,just complicated

Teachers like to give students those questions

[wanton sky]

To get these solutions, I needed to find and solve a quartic equation, and it was way beyond what could be reasonable expected at school. If I wanted ti ask this qustion, I would have to put in structure to guide students as to an approach to take.

[pseudo umbra]

What answer did you get?. You wrote quite a lot so I cants see it clearly

[wanton sky]

There are two possibilities:

$x = 3 + \sqrt{109} - \sqrt{82-6\sqrt{109}} \approx 9.0405...$ cm.

or

$x = 3 + \sqrt{109} + \sqrt{82-6\sqrt{109}} \approx 17.8401...$ cm.