Maths Game (1 Viewer)

[Kabuzo01]

Well since the SC is coming up I reckon we should have a bit of fun.... with MATHS!!! -.-" Anyways if you have a maths question that you dont understand or want to see if anyone can solve it, post it here, even the hardest of the hardest maths questions will be interesting to solve.

And if you get any correct, or the person cant answer your question (which has a proper answer) you get a point. (note that these points are not useable from any reasons) o.o

Rules

Who gets the answer correct first gets the point and also make sure if you edit your answer no one has posted it already.

By that I mean say Person 1 posted an incorrect answer
Person 2 posts the correct answer
Person 1 sees this and changes the answer to be the same as Person 2.

This can be a mistake if someone accidently edits their post to change wordings or something. PM me if you dont understand.


All Questions Must Be SC and Lower leveled or general knowledge type of question, not 4 unit maths >_> And the person who asks the question must know the answer beforehand.




~I'm going to edit more after the SC exams, if anyone wants to include anything that should be useful in the rules just PM me or comment in this thread.

 

[supercalamari]

Maths? Fun? Uhhh-nuhhh.

 

[DALiE. (:]

Math and fun should not be used in the same sentence. =D

 

[Kabuzo01]

Math and fun should not be used in the same sentence. =D

You just used it =D Plus math is fun.... that reminds me... mathisfun.com (tanks game) is blocked (school intranet) =O

 

[lolman12567]

taken from baulko year 10 trial, quite a simple but awesome question.

In the Diagram below ABCD and DEFG are squares,

show that CF² + BE² = 3(AD² + DE²)

 

[Kabuzo01]

taken from baulko year 10 trial, quite a simple but awesome question.

In the Diagram below ABCD and DEFG are squares,

show that CF² + BE² = 3(AD² + DE²)

http://img171.imageshack.us/my.php?image=squarzdb9.png

I done it using the subsitutional method, it was easy but its gonna be a pain in the arse to type it up.

To show that CF² + BE² = 3(AD² + DE²) You will need to subsitute the squares.
For mine i made AB= 2 and GD = 3

CG= 1 and GF = 3 So using pythag theorm FC²= 10
AB = 2 and B somewhere between FE to make a even rectangle with the sides A B and E with the total length of 5 and using pythag it equals to 29

So 10 + 29 = 39


Now AD²= 4
And DE²= 9
AD² + DE² = 13 x 3 = 39

Therefore CF² + BE² = 3(AD² + DE²)

One point for me ^_^

 

[DALiE. (:]

You just used it =D Plus math is fun.... that reminds me... mathisfun.com (tanks game) is blocked (school intranet) =O

True. But there are exceptions to every rule.

 

[Kabuzo01]

True. But there are exceptions to every rule.

=D And rules are made to be broken ^_^

 

[lyounamu]

I done it using the subsitutional method, it was easy but its gonna be a pain in the arse to type it up.

To show that CF² + BE² = 3(AD² + DE²) You will need to subsitute the squares.
For mine i made AB= 2 and GD = 3

CG= 1 and GF = 3 So using pythag theorm FC²= 10
AB = 2 and B somewhere between FE to make a even rectangle with the sides A B and E with the total length of 5 and using pythag it equals to 29

So 10 + 29 = 39


Now AD²= 4
And DE²= 9
AD² + DE² = 13 x 3 = 39

Therefore CF² + BE² = 3(AD² + DE²)

One point for me ^_^

Ah, I am not sure if you are allowed to do so.

You cannot make a definite judgement to what the values can be. If you really wanted to use the substitution method, I suggest you using values such as x or y or z.

 

[Kabuzo01]

Ah, I am not sure if you are allowed to do so.

You cannot make a definite judgement to what the values can be. If you really wanted to use the substitution method, I suggest you using values such as x or y or z.

However it proved that they are all equaled ^_^

 

[lyounamu]

However it proved that they are all equaled ^_^

Yes, it did. But it would be far much better if you proved it the other way.

You cannot just "assume" that they are equal, unless it's a mathematical induction question. If you go to senior level of mathematics, you will potentially lose marks because of that, as a matter of fact, I am pretty sure you will lose marks.

 

[Kabuzo01]

Yes, it did. But it would be far much better if you proved it the other way.

You cannot just "assume" that they are equal, unless it's a mathematical induction question. If you go to senior level of mathematics, you will potentially lose marks because of that, as a matter of fact, I am pretty sure you will lose marks.

Probably =D But this is year 10 ^_^ =P I've used that method many times before and I always got full marks for it. But he asked for the answer not the working out =D But if you trying to give me advice about the risks in Senior levels, thanks ^_^

 

[lyounamu]

Probably =D But this is year 10 ^_^ =P I've used that method many times before and I always got full marks for it. But he asked for the answer not the working out =D But if you trying to give me advice about the risks in Senior levels, thanks ^_^

okay. May be it was wrong to have invaded you with all those advices.

 

[Kabuzo01]

okay. May be it was wrong to have invaded you with all those advices.

=P Iono lolz

 

[pwoh]

Using x, y, z values like lyounamu suggested:

Prove CF² + BE² = 3(x² + y²)

x = BA = AD = CD (sides of a square are equal)
y = DE = GF = GD (sides of a square are equal)
CG = z = y - x
x = y - z
y = z + x

Pythagoras' Theorem:
z² + y² = CF² ________(1)
x² + (x + y)² = BE²____(2)
x² + (2x + z)² = BE²___(3)

(1) + (2) = CF² + BE²
= z² + y² + x² + (x + y)²___(4)

Substitute z = y - x into (4)
CF² + BE² =(y - x)² + y² + x² + (x + y)²
= y² -2xy + x² + y² + x² + x² + 2xy + y²
= 3x² + 3y²

Therefore CF² + BE² = 3(x² + y²)


(It's a bit messy... )

A question from our yearly:

If sin A = 5/7 and A is obtuse, find the exact value of tan A.

 

[lyounamu]

Using x, y, z values like lyounamu suggested:

Prove CF² + BE² = 3(x² + y²)

x = BA = AD = CD (sides of a square are equal)
y = DE = GF = GD (sides of a square are equal)
CG = z = y - x
x = y - z
y = z + x

Pythagoras' Theorem:
z² + y² = CF² ________(1)
x² + (x + y)² = BE²____(2)
x² + (2x + z)² = BE²___(3)

(1) + (2) = CF² + BE²
= z² + y² + x² + (x + y)²___(4)

Substitute z = y - x into (4)
CF² + BE² =(y - x)² + y² + x² + (x + y)²
= y² -2xy + x² + y² + x² + x² + 2xy + y²
= 3x² + 3y²

Therefore CF² + BE² = 3(x² + y²)


(It's a bit messy... )

Well done (I didn't read into great details but your approach seems right).

 

[Kabuzo01]

A question from our yearly:

If sin A = 5/7 and A is obtuse, find the exact value of tan A.[/quote]


Shit... year 11 work, sin on a obtuse triangle o.o I give up =D Is that 2 unit? Looks like general maths lolz

 

[lyounamu]

A question from our yearly:

If sin A = 5/7 and A is obtuse, find the exact value of tan A.

Shit... year 11 work, sin on a obtuse triangle o.o I give up =D Is that 2 unit? Looks like general maths lolz[/quote]

this is 2 unit level maths from yr 11. But it can be perceived as a harder yr 10 maths too.

tanA is negative when it goes to the 2nd quadrant (i.e. obtuse angle basically).

So since sine = opp/hyp

adjacent = sqrt(24) (because 7^2 = x^2 + 5^2)

tanA = opp/adj = 5/-sqrt(24) = -5sqrt(24)/24

 

[pwoh]

One thing I don't understand...how come we can use Pythagoras' Theorem there if it doesn't say there is a right angle? (That does use Pythagoras right?)

I got that question right (apart from the negative which I forgot >O) but that's only because I had looked up how to do it beforehand...

 

[lyounamu]

Good question. You can use the pythagoras' theorem because it's just easiest way to approach the question. I really don't see how I can approach this without using that method.

In the question, you are only interested in the value of the tanx. Since the answer you get from the question conforms to the value of the sine (which was initially given), your answer is valid, hence your working out is valid (but not necessarily the case).