Euclid eWorkshop Questions

[whoareyou]

After they defined the opposite and adjacent leg to be a and b, how did they determine the rest? I don't understand why subtracting the length of the legs gives you the side length of the smaller square ... is there a property for that?

The second I'm guessing is area.

Ok yah, everything after the a-b = 3 I get.

Can someone explain how they got a-b = 3?

[Brightguy]

If you subtract the length of triangle's long leg from its short leg you get the side length of the inner cube.

[whoareyou]

Could you link me to a website discussing the proof?

Also, do you think this question could have been solved without knowing that property?

[Zren]

I think they typoed a & b. As:
a - b = 3
a = 3 + b or b = a - 3
Which makes no sense as a = opp, and b = adj.

All 4 triangles are the same. So if you look at side b (adj), you notice that part of it is made of of a 'side a (opp)' of another of the similar triangles + the length of the side of the small square.

b = a + (side of square)
b = a + x
x = (side of square)
(area of square) = 9 = x^2
x = sqrt(9)
x = +-3 (negative is ignored because you can't have negative length)
b = a + 3


... Thus, typo?

[Brightguy]

whoareyou @ Sun Apr 01, 2012 12:08 am wrote:

Could you link me to a website discussing the proof?

Just look at the diagram. The long leg is made up of two line segments.

whoareyou @ Sun Apr 01, 2012 12:08 am wrote:

Also, do you think this question could have been solved without knowing that property?

Yes, you can also use the Pythagorean Theorem to get another relation between a and b.

Zren @ Sun Apr 01, 2012 12:16 am wrote:

I think they typoed a & b. As:
a - b = 3
a = 3 + b or b = a - 3
Which makes no sense as a = opp, and b = adj.

a is the longer side, so naturally a − b will be positive...

[Zren]

Brightguy @ Sun Apr 01, 2012 12:27 am wrote:

whoareyou @ Sun Apr 01, 2012 12:08 am wrote:

Could you link me to a website discussing the proof?

Just look at the diagram. The long leg is made up of two line segments.

whoareyou @ Sun Apr 01, 2012 12:08 am wrote:

Also, do you think this question could have been solved without knowing that property?

Yes, you can also use the Pythagorean Theorem to get another relation between a and b.

Zren @ Sun Apr 01, 2012 12:16 am wrote:

I think they typoed a & b. As:
a - b = 3
a = 3 + b or b = a - 3
Which makes no sense as a = opp, and b = adj.

a is the longer side, so naturally a − b will be positive...

Nevermind. I somehow managed to put the omega symbol on the wrong side in my sketch. FML. Btw. Tipsy yet?

[whoareyou]

I have a math question, but I didn't want to make a new topic.

So anyways, if I have the equation (-2-x)^2 = (4-x)^2, why can't I just take the square root of both sides and solve (it doesn't work btw)?

[bl0ckeduser]

whoareyou @ Thu May 24, 2012 3:55 pm wrote:

I have a math question, but I didn't want to make a new topic.

So anyways, if I have the equation (-2-x)^2 = (4-x)^2, why can't I just take the square root of both sides and solve (it doesn't work btw)?

I'm not a math expert, but this is what I remember from my algebra class:

There are always two square roots, one positive, and one negative (for example, 3^2 = (-3)^2 = 9). The square root operation is defined as giving the positive one. By taking the square root of both sides you are losing information, because you are ignoring the negative square roots and only keeping the positive ones.
What you can do is say |-2-x| = |4-x|, or say (-2-x)^2 - (4-x)^2 = 0, expand, and solve the resulting equation, which will be at worse a quadratic.

[y4y]

whoareyou @ Thu May 24, 2012 3:55 pm wrote:

I have a math question, but I didn't want to make a new topic.

So anyways, if I have the equation (-2-x)^2 = (4-x)^2, why can't I just take the square root of both sides and solve (it doesn't work btw)?

multiply both sides by -1

also was wondering where Euclid threads were. lol wish i saw it before maybe i might have actually would have attempted preparation....

[whoareyou]

y4y @ Thu May 24, 2012 7:23 pm wrote:

whoareyou @ Thu May 24, 2012 3:55 pm wrote:

I have a math question, but I didn't want to make a new topic.

So anyways, if I have the equation (-2-x)^2 = (4-x)^2, why can't I just take the square root of both sides and solve (it doesn't work btw)?

multiply both sides by -1

also was wondering where Euclid threads were. lol wish i saw it before maybe i might have actually would have attempted preparation....

Why?

[y4y]

whoareyou @ Thu May 24, 2012 7:25 pm wrote:

y4y @ Thu May 24, 2012 7:23 pm wrote:

whoareyou @ Thu May 24, 2012 3:55 pm wrote:

I have a math question, but I didn't want to make a new topic.

So anyways, if I have the equation (-2-x)^2 = (4-x)^2, why can't I just take the square root of both sides and solve (it doesn't work btw)?

multiply both sides by -1

also was wondering where Euclid threads were. lol wish i saw it before maybe i might have actually would have attempted preparation....

Why?

seems i was wrong...forget what i said....

[y4y]

bl0ckeduser seems to be right.

if you were to take the square root you would have to consider 4 equations

-2-x=4-x
-2-x=x-4
2-x=x-4
2-x=4-x

that's what i got...

the second and third are easily solvable.

[Zren]

It does work.
Edit: Goddamn CompSci and it's lack of unicode... ? is the + or - symbol.

sqrt(9) = ?3

(-2-x)^2 = (4-x)^2
sqrt((-2-x)^2) = sqrt((4-x)^2)
?(-2-x)= ?(4-x)

4 Possibilities:
+(-2-x) = +(4-x)
+(-2-x) = -(4-x)
-(-2-x) = +(4-x)
-(-2-x) = -(4-x)

Full Answers below warning.

This is probably what you tried.
+(-2-x) = +(4-x)
-2-x = 4-x
-x + x = 4 + 2
0x = 6
Cancelled out. This also happens with the - and - combo. Truthfully, I'm not entirely sure the exact terminology for when this happens.

Using the opposite + and - (or - and +) possibility lead to:
+(-2-x) = -(4-x)
-2 - x = -4 + x
-x - x = -4 + 2
-2x = -2
2x = 2
x = 1

You should really be expanding both sides though. It's way faster.
(-2-x)^2 = (4-x)^2
(-2-x)(-2-x) = (4-x)(4-x)
4 + 4x + x^2 = 16 - 8x + x^2
12x = 12
x = 1

[crossley7]

So starting with (-2-x)^2 = (4-x)^2

get rid of the negative in the left side by factoring and then squaring the -1 to become 1. Now the cleanest solution is to expand simplify and then either factor or use the quadratic formula depending on what you are left with. It is a lot better than dealing with the cases that you get by taking the square root immediately.


I can tell you from looking at it that x=1 should be the answer. Also, in this case, taking the square root will work if you get the negative out of the brackets before taking the root.