I call you a genius if... EDIT: SOLUTION now up!

[MichaelD]

Pythagorus says "no."

 

[Vertimus]

Originally posted by: MichaelD
Pythagorus says "no."

Why Not?

 

[MartyMcFly3]

Originally posted by: Vertimus

Originally posted by: MartyMcFly3
42

Use the refresh button more often

Its my way of saying im too lazy to do it.

 

666

 

[MichaelD]

Originally posted by: Vertimus

Originally posted by: MichaelD
Pythagorus says "no."

Why Not?

B/C, he's Pythagorus. He invented his theorum. He thinks this is easy sh{t. He doesn't want to help you.

 

[screw3d]

OMGHI2U!

 

[TallBill]

42

 

[Ausm]

Originally posted by: Vertimus

Originally posted by: bootymac
A = B * H / 2

That's all I'm doing for you

What's B, and H?

Base

Height


Sysadmin

 

[Alchemist99]

>.999...

 

[TGregg]

Originally posted by: MichaelD
Pythagorus says "no."

Doesn't he just talk about right triangles?

 

[geno]

this would be easier if <BFC was 90*

 

[Pepsi90919]

Originally posted by: MichaelD

Originally posted by: Vertimus

Originally posted by: MichaelD
Pythagorus says "no."

Why Not?

B/C, he's Pythagorus. He invented his theorum. He thinks this is easy sh{t. He doesn't want to help you.

lol

 

[blahblah99]

Do your own homework problem, or pay me $5 to do it.

 

[TuxDave]

If you solved it, feel free to post the answer in a couple of minutes.

 

[eelw]

Originally posted by: Vertimus

Originally posted by: purbeast0
without angles this problem is not solvable ...

Wrong. I solved it, although it took me 35 minutes.

No, I would have to agree with purbeast, without angles, you can't solve this. Unless you're assuming angles BFC and BEC are 90 degrees.

 

[Vertimus]

Originally posted by: eelw

Originally posted by: Vertimus

Originally posted by: purbeast0
without angles this problem is not solvable ...

Wrong. I solved it, although it took me 35 minutes.

No, I would have to agree with purbeast, without angles, you can't solve this. Unless you're assuming angles BFC and BEC are 90 degrees.

I'm willing to bet $50 dollars via paypal you are wrong. (BTW, please don't bet, since I already solved it.)

 

[Vertimus]

Originally posted by: blahblah99
Do your own homework problem, or pay me $5 to do it.

Read the post. I already solve it.

 

[TuxDave]

Originally posted by: eelw

Originally posted by: Vertimus

Originally posted by: purbeast0
without angles this problem is not solvable ...

Wrong. I solved it, although it took me 35 minutes.

No, I would have to agree with purbeast, without angles, you can't solve this. Unless you're assuming angles BFC and BEC are 90 degrees.

It might be solvable. I ended up with only one unknown and I haven't used the length 20 segment yet... but I can't figure out to intergrate that 20 into my information.

 

[43st]

Originally posted by: TuxDave

Originally posted by: eelw

Originally posted by: Vertimus

Originally posted by: purbeast0
without angles this problem is not solvable ...

Wrong. I solved it, although it took me 35 minutes.

No, I would have to agree with purbeast, without angles, you can't solve this. Unless you're assuming angles BFC and BEC are 90 degrees.

It might be solvable. I ended up with only one unknown and I haven't used the length 20 segment yet... but I can't figure out to intergrate that 20 into my information.

Same here. The 20 length has no relationship to anything. If we knew the bottom was a straight line it would help a bit but we don't know that. The problem is missing information.

 

[simms]

Use sine/cosine laws to do it. I would, but too lazy. No incentive.

 

[Brutuskend]

Here's your answer!

 

[upsciLLion]

I bet some tricky usage of Heron's formula for triangular area would lead to a solution.

 

[Legendary]

If BE and FC bisect the angles this is easy. Or if AD forms a right angle with BC. Or if BE forms a right angle with AC.
If they don't, they I don't have any idea. None of this information is given.

 

[bonkers325]

the problem is deceiving you. it is not drawn correctly, so the inscribed triangles look different, but they're actually the same. the have common bases and one common side, which makes the 3rd side equivalent.

 

[43st]

Are A-B, B-C, C-D straight lines?