# Who knoes the answer of this:

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• Aug 1st 2010, 03:48 AM
Wissem
Who knoes the answer of this:
In the Hotel lodge, your nightly hotel price gets you a bedroom and a bathroom. Each room is perfectly square. If you take the length of one of the walls of the bedroom, and subtract the length of one of the walls of the bathroom, you'll find that the difference is 9 metres. Now, if you multiply the length of the bedroom wall by the length of the bathroom wall, you'll find that the product is 12 square metres.

What is the total area, in square metres, of the bedroom plus the bathroom? Round to the nearest whole number, if necessary.

• Aug 1st 2010, 04:03 AM
Quote:

Originally Posted by Wissem
In the Hotel lodge, your nightly hotel price gets you a bedroom and a bathroom. Each room is perfectly square. If you take the length of one of the walls of the bedroom, and subtract the length of one of the walls of the bathroom, you'll find that the difference is 9 metres. Now, if you multiply the length of the bedroom wall by the length of the bathroom wall, you'll find that the product is 12 square metres.

What is the total area, in square metres, of the bedroom plus the bathroom? Round to the nearest whole number, if necessary.

You need to write a pair of equations for the 2 lengths,
then get that down to a single length.

You can use x and y for this.
Suppose x is the length of the larger wall.

x-y=9
xy=12

Now you can write one variable in terms of the other.
That gets you down to one variable,
because the x is the same in both lines.
So is the y.

x=9+y
(9+y)(y)=12

Can you continue?

Finally, you are asked for $\displaystyle x^2+y^2$

so you can go another route

$\displaystyle (x-y)^2=x^2-2xy+y^2=9^2$

then use xy=12
• Aug 1st 2010, 04:04 AM
Unknown008
Well, I think that's a simple problem of simultaneous equations...

Let x be the length of one side of the bathroom.
Let y be the length of one side of the bedroom.

So,

Quote:

If you take the length of one of the walls of the bedroom, and subtract the length of one of the walls of the bathroom, you'll find that the difference is 9 metres.
This means $\displaystyle y - x = 9$

Quote:

Now, if you multiply the length of the bedroom wall by the length of the bathroom wall, you'll find that the product is 12 square metres.
This means $\displaystyle yx = 12$

Can you solve it now?

EDIT: A little too late... but btw, is this really in the correct section? I mean, it's in the University section, while I'm pretty sure it'll be better in the pre-university section...
• Aug 1st 2010, 04:42 AM
Wissem
I did something like this but I don't understand this (9+y)(y)=12
?from where do we got the (9+y)(y)=12?
So 12=x?
• Aug 1st 2010, 04:45 AM
Wissem

so you can go another route

then use xy=12

---

and I don't understand this.

You may think that I'm stupid but I'm 14 :P

• Aug 1st 2010, 04:55 AM
Quote:

Originally Posted by Wissem
I did something like this but I don't understand this (9+y)(y)=12
?from where do we got the (9+y)(y)=12?
So 12=x?

This is the longer way to do it!

The easiest way, if you have been learning about multiplying out factors is as follows..

The area of one bedroom is $\displaystyle x^2$ square units

and the area of the other bedroom is $\displaystyle y^2$ square units.

If we let the bigger room have side=x (we could let the bigger one be y of course, it's up to us),
then x-y=9.

Now, your question is asking us for $\displaystyle x^2+y^2$

because that is the area of both bedrooms combined.

What happens when we multiply out $\displaystyle (x-y)^2$

knowing that it must equal $\displaystyle 9^2=81\ ?$

This is the part you must become proficient at....

$\displaystyle (x-y)^2=(x-y)(x-y)=x(x-y)-y(x-y)=x(x)+x(-y)-y(x)-y(-y)$

$\displaystyle =x^2-xy-xy+y^2=x^2+y^2-2xy$

This is 81 and xy=12.

therefore

$\displaystyle x^2+y^2-2xy=81$

$\displaystyle x^2+y^2-2(12)=81$

$\displaystyle x^2+y^2-2(12)+2(12)=81+2(12)$

$\displaystyle x^2+y^2=81+24=105$ square units.

(Note: you're very young and don't take any notice of anyone who might think anything wrong of you)
• Aug 1st 2010, 05:34 AM
Wissem
Woowh, very difficult!

• Aug 1st 2010, 06:01 AM
Quote:

Originally Posted by Wissem
Woowh, very difficult!

So the andwer is 105m2?

The main thing is to be able to do the calculations.
If you can do them, then you will easily be able to get the answer.

What parts of the preceding are difficult
and have you covered those details in class?

Remember to think in terms of numbers.
For instance

$\displaystyle 7-4=3$

$\displaystyle 7(4)=28$

$\displaystyle (7-4)^2=3^2=9$

but it's also $\displaystyle (7-4)(7-4)=7(7-4)+(-4)(7-4)=7(7)+7(-4)-4(7)-4(-4)$

$\displaystyle =49-28-28+16=49+16-2(28)=65-56=9$

It's the same procedure if you put other numbers there in place of the 7 and 4,
it works for all numbers so we can put x and y there.

Are you familiar with (-)(-)=+ and (-)(+)=- and so on?
• Aug 1st 2010, 06:06 AM
Wissem
Quote:

Are you familiar with (-)(-)=+ and (-)(+)=- and so on?

Yes I know that. :) but it stills difficult. In which class do you learn this?
Because After the summer, I'm n the 3th class. so 3 years after the last year of the primair education.
I hope you'll understand me ^^
• Aug 1st 2010, 06:14 AM
Wissem
I wrote it on paper and I understand it now.
But I still don't understand 1 thing:
First we have: 7-4=3
7(4)=28
(7-4)2 =32= 9

Q: Why did we do this? that 2?? (7-4)2.
Where did it come from?
• Aug 1st 2010, 06:19 AM
Wissem
(x-y)^2=(x-y)(x-y)=x(x-y)-y(x-y)=x(x)+x(-y)-y(x)-y(-y)

How do we know the x ?
• Aug 1st 2010, 06:26 AM
Quote:

Originally Posted by Wissem
Yes I know that. :) but it stills difficult. In which class do you learn this?
Because After the summer, I'm n the 3th class. so 3 years after the last year of the primair education.
I hope you'll understand me ^^

It should be introduced at some stage of your coming year,
probably early on.

The foundation to all of this is to understand that
the reason we multiply one side by another to get the area of a room
is because that's the fastest way to count the number of "squares" inside it
that are 1 unit by 1 unit in measurement,
as if the floor was tiled by squares with all 4 sides 1 unit long.

Hence if a room is square, the number of square units inside is the square of a side length.
If a room is rectangular, the number of square units inside is obtained by multiplying the
lengths of the 2 perpendicular sides.

Also, understanding how to multiply out factors then helps you to answer your question.
You just need to learn and practice with that.

If you draw 5 rows of 5 dots, there are 5(5)=25 dots.

Now draw a square around the top left 3 rows of 3.
This leaves you with 2 rows of 2 in the bottom right.
The rest is 2 rows of 3 and 3 rows of 2.

Hence (5)(5)=(3)(3)+(2)(2)+(3)(2)+(3)(2)

Factorising is based on that

(5)(5)=(3+2)(3+2)=3(3+2)+2(3+2)=3(3)+3(2)+2(3)+2(2 )

Then (x+y)(x+y) or (x-y)(x-y) is multiplied out in the same way.
• Aug 1st 2010, 06:35 AM
Wissem
wowh ^^ Very difficult English. I must translate it to Dutch so I'll understand it :P
It's not for school or something, I found it on the web.

I'm understanding it now.
x(x)=81 and y(y)= 24
Is that good?
• Aug 1st 2010, 06:38 AM
Quote:

Originally Posted by Wissem
I wrote it on paper and I understand it now.
But I still don't understand 1 thing:
First we have: 7-4=3
7(4)=28
(7-4)2 =32= 9

Q: Why did we do this? that 2?? (7-4)2.
Where did it come from?

This is just an example to show you that
if we just write 7-4=3, then (7-4)(7-4)=3(3)=9.

But when you have (x-y)(x-y) we can't do that, unless we know x and y.

so, it's just to show that if you multiply out all the terms in the right way, you get
an alternative version of the correct answer.

It was just for illustration of the techniques you will be learning.
• Aug 1st 2010, 06:43 AM
Quote:

Originally Posted by Wissem
(x-y)^2=(x-y)(x-y)=x(x-y)-y(x-y)=x(x)+x(-y)-y(x)-y(-y)

How do we know the x ?

In your example, the funny thing is... we didn't need to find out what x is! or y either!
By multiplying out the factors, we found the combined area of both rooms,
using the 2 clues given, without ever needing to know x or y.

If you calculate x or y's value, you will need to know the "quadratic formula"..

$\displaystyle x-y=9$

$\displaystyle xy=12$

$\displaystyle x-y+y=9+y$

$\displaystyle x=9+y$

$\displaystyle xy=(9+y)y$

$\displaystyle (9+y)y=9y+y^2=12$

$\displaystyle y^2+9y-12=0$

$\displaystyle ay^2+by+c=0\ \Rightarrow\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

I don't know if you've covered that.

Once you know what y is, then you find x.
Seems morre complex and you may not be that far yet.
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