\(ACEG\) is a rectangle. If segment \( \color {blue}{BE}\) is \( 30\), segment \(\color{red}{CG}\) is \(40\), segment \(\color{green}{DF}\) is \(15\) and \(\angle FDE = 90^\circ\). Find \(CE\)

*This problem has been solved*

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI'm sorry for my previous incorrect comment. The approach I suggested leads to a trigonometric equation that I don't see an easy way to solve. Same for just using Pythagorean identities:

\[CD = 15\frac{w}{\sqrt{40^2-w^2}} = w-15\frac{w}{\sqrt{30^2-w^2}}\]

Wolfram Alpha has the same numerical solution for both, though: just under 16 (15.9876490085686).

Log in to reply

So the answer is approach to \(16\) ?

So what formula we use ? Both Pythagorean and Trigonometry ?

Log in to reply

You guys rocked it.! Nice team work..& hats off to ur efforts..!! Me & my friends couldn't solve it.. I really need to learn a lot from people like you guys...congrats.

Log in to reply

LOL thx

Log in to reply

Jason, where did you find this problem? Are you sure there's the sort of clean solution we've been looking for?

Log in to reply

I find the \(SOLUTION\) !!!!

I just need someone to simplify an expression

Use Pythagoras and the rule of similarity

Log in to reply

See told ya pythagoras would come handy

Log in to reply

Log in to reply

It's from a piece of paper that have been torn off by someone. I found it on school.

Log in to reply

Hello Jason, thanks for inviting to this problem XD. Somehow I got the answer for GF/FE . Trying to solve further.

Log in to reply

Okay, LOL

Log in to reply

Yeah, I think I got the way to solve. I have got an equation in x, but a lengthy one, I have to solve that. An you help me with that.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

I actually know the answer but dunno how to get it. \(CE \approx 16\)

Log in to reply

Log in to reply

Log in to reply

My algebra level = 2

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

1)450=BC×FE

2)600=CF×EG

3)4/3=FG/FE

4)600=BC×FG

5)BC/CF=EG/FG

6)1200=GE×BC

7)8/3=GE/FE

8)2CF=BC

9)2FG=GE

10)225=CF×FE

11)300=CF×FG

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Rishabh Cool @Calvin Lin Help us XD

KLog in to reply

@Paola Ramírez

AlsoLog in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

PS:Maybe @Nihar Mahajan could be the right guy to consult here as his geometry skills are very good...

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Use coordinate geometry

Log in to reply

Is there any other solution

Log in to reply

I think we should invite some seniors at brilliant to look for a solution to this problem..

Log in to reply

Can u please post your solution ? It would be very nice because this problem is making very lengthy and tedious eq. So plz help us.

Log in to reply

@ahmad saad

Log in to reply

@Nihar Mahajan @Svatejas Shivakumar Help usssss

Log in to reply

I have found the solution, use the rule of Pythagoras and Similarity

I need someone to FIND \(x\)this expression below ( \(x = CE\))

\(\Huge 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x \)

I'll explain how to get the expression above after someone find x

Log in to reply

Okay, that says that the area of the whole rectangle = area of bottom 15 of rectangle (15x) plus that same 15x area divided by the height where the line segment of length 30 meets the left side of the rectangle. That can't be right. In any case, we have good equations in x that "just need to be solved/simplified". The problem is that it is not at all clear that they can be.

Edit: you've changed the equation since I wrote my comment. Anyway, to spend more energy on this I would want some stronger reason to believe that there is a simple solution to be found.

Log in to reply

I'm serious, input \(x= 15.9876\), the expression above

I find the expression but can't find \(x\)

Log in to reply

Assume that : \( CE = x \), \( DE = y\)

That's mean \( CD = x-y\)

Using Pythagoras :

\( BC = \sqrt{900 - x^2}\)

\( EG = \sqrt{1600 - x^2}\)

Using Similiar Triangles :

\( \LARGE \frac{FD}{BC} = \frac{DE}{CE} \\ \LARGE \frac{15}{\sqrt{900-x^2}} = \frac{y}{x} \\ \LARGE y =\frac{15x}{\sqrt{900-x^2}}\)

\(\LARGE \frac{FD}{EG} = \frac{CD}{CE} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-y}{x} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-\frac{15x}{\sqrt{900-x^2}}}{x}\)

\(\LARGE 15x = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) \)

\(\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x \)

\(\LARGE \text{SOMEONE FIND}\space \Huge x \space \LARGE \text{PLZZZZ !!!!!! }\)

Log in to reply

@Mark C @Abhay Kumar @Ashish Siva @Rishabh Tiwari @Rohit Udaiwal

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Mystery Solved ;)

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

From the expression above, prove that \(x= 15.9876...\) ?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Jon Haussmann maybe you can help :)

Yep, this queetion may be wrong but lets see if we can get some easier method. Oh! I forgotLog in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Because it might look simple but actually it's hard ( took us 3 day to solve LOL)

Log in to reply

Log in to reply

Log in to reply

Log in to reply

click here

Done,Log in to reply

Log in to reply

Log in to reply

@Abhiram Rao @Abhay Tiwari help plz

Log in to reply

Use Pythagoras Theorem. But the solution would be a very complex one.

Log in to reply

So we have found the answer and how to get it

Should i delete this discussion ?

Log in to reply

Nooooo, let it be there, we have worked very hard on this.

Log in to reply

Okay XD

Log in to reply

Log in to reply

Guys, i make a problem based on this discussion :)

Log in to reply

Unfortunately u had made a mistake so the answer is wrong :(

Log in to reply

Wait what !!!

Log in to reply

Log in to reply

Where ?

Log in to reply

Your answer is wrong.Mistake founder-Abhay kumar.

Log in to reply

AHHHHHHHHHHHHHHHHHH

Log in to reply

I'm confused by all the many comments. Has someone found a solution that simplifies, or are we still stuck with equations we need a computer to solve?

Log in to reply

We solved it! With a little help of W/A

Log in to reply

But we'd (I'd) already done that two days ago. Sort by top comments & the first is mine from two days ago. I still wonder if there is a W/A-free way to do it.

Log in to reply

Log in to reply

I loved the way you solved the problem and thx.....guys for such a nice sol. Btw nice prob . Rishabh

Log in to reply

Oh hi pranay. Thanks for your encouragement.

Log in to reply