Question

if the perimeter of a rectangle is p and its diagonal is d, the difference between the lenght and width of the rectangle is?

Answer 1

Let x and y equal the lengths of the two sides of the rectangle. What is the formula for the perimeter p based on x and y . If the diameter is d, what would be the formula for that in terms of x and y.

Answer 2

the formula????

Answer 3

The formula for perimeter is 2(l + b)
So, basically, 2(l + b) can be said as the perimeter.
Now, you'd know the Pythagorean theorem. It says that a^2 + b^2 = c^2
We can say that d^2 = l^2 + b^2

Continue now.

Answer 4

Perimeter p = 2x + 2y. The perimeter is the sum of the lengths of the 4 sides.

Answer 5

i am confused so if the formula is p = 2x + 2y, then what do i do?

Answer 6

As ParthKohli has mentioned,
\[d ^{2} = l ^{2} + b ^{2}\] and
\[p = 2l + 2b\]
Can you find l - b now?

Answer 7

yeay thats much simpler thanks!

Answer 8

Actually, even I'm confused now, how do you proceed?

Answer 9

p = 2(l + b)
= -2(-l - b)
-l - b = -p/2

Answer 10

Now what?

Answer 11

oh wait i thought you use the Pythagorean theorem to solve the rest??

Answer 12

this problem is hopeless. i'm going to kill my teacher

Answer 13

I'm not so sure ... why don't we take the advice of @ParthKohli?

Answer 14

No please don't kill your teacher. lol

are you familiar with pythagoras theorem?

Answer 15

do you mean the Pythagorean theorem???

Answer 16

Yes.

Answer 17

yes

Answer 18

i am

Answer 19

EVERYONE knows Pythagorean theorem .

Answer 20

Did your teacher give you any other information like what d and p are?

Answer 21

sadly no,

Answer 22

Wow, I have been trying this all different ways and haven't gotten anywhere with it. Maybe we should ask @FoolForMath to step in and help.

Answer 23

i am giving up. thanks for trying help out you guys

Answer 24

Sorry. That one is a stumper.

Answer 25

@Zarkon, can you help?

Answer 26

Lol I'm not sure what I have to do here

Answer 27

@ParthKohli We can't seem to get this, @tim538 seems to have given up but it bugs me that I can't get it.

Answer 28

Maybe the difference of the lengths and width is l - b lolz

Answer 29

i'll show you link, it number 4

i don't know if this helps but you can see original problem for yourself.

Answer 30

I'm pretty sure about that one

Answer 31

2(l+b)=p => (l+b)=p/2 => l^2 +b^2+2lb= p^2/4 => 2lb= p^2/4-d^2

(l-b)^2= l^2+b^2-2lb => \( (l-b)= \sqrt{2d^2-p^2/4} \)

Answer 32

i said 3 instead of 4

Answer 33

@FoolForMath you got it that easily!!!

Answer 34

;)

Answer 35

From the formula for the perimeter\[2(x+y)=p\]Rearrange and square both sides\[(x+y)=p/2\]\[(x+y)^2=\frac {p^2}{2^2}\]\[x^2 +y^2+2xy= p^2/4\]Rearrange again to solve for 2xy\[2xy=\frac {p^2}{4}-(x^2+y^2)\]Remember and substitute \[x^2 + y^2=d^2\]\[2xy=\frac{p^2}{4}-d^2\]\[(x-y)^2= x^2+y^2-2xy\]\[(x-y)^2=d^2-\left(\frac{ p^2}{4}-d^2 \right)=-\frac {p^2}{4}\]\[(x-y)= \sqrt{-p^2/4}\]

Answer 36

@FoolForMath, does the d^2 term drop at the end?

Answer 37

*doesn't

Answer 38

I don't know the pythagorean theorem. Trolol

Answer 39

\[(x-y)^2=d^2-\left(\frac{ p^2}{4}-d^2 \right)=d^2-\frac {p^2} 4+d^2 =2d^2 -\frac {p^2}{4} \]

Answer 40

@FoolForMath, yes, math mistake. Thanks for clearing that up.