Relationship between Area of Square and Rectangle

Togechu64 · #1 $Old$ November 3rd, 2009, 08:18 PM

(a) A rectangle has the same area as a square. The long side of the rectangle is four times the length of the short side. The perimeter of the square has length 160 mm. How much longer is the perimeter of the rectangle than the perimeter of the square?
(b) Same question but this time the perimeter of the square is L mm and the long side of the rectangle is x times as long as the short side.

I've already gotten (a) correct, but no matter how many times I try (b) I haven't gotten the right answer.

ukorov · #2 $Old$ November 3rd, 2009, 09:39 PM

in terms of what is that right answer supposed to be?

Togechu64 · #3 $Old$ November 3rd, 2009, 09:42 PM

But in part B, the length is "x" times the length of the width, not 4 times the length of the width like in part a.

pflo · #4 $Old$ November 3rd, 2009, 09:44 PM

ukorov -
Your solution states the length of the rectanble is 4x when the width is x. That means the length must be 4 times the width. But that is not what the problem states in part b.

ukorov · #5 $Old$ November 3rd, 2009, 09:46 PM

Quote:

Originally Posted by pflo $View Post$

ukorov -
Your solution states the length of the rectanble is 4x when the width is x. That means the length must be 4 times the width. But that is not what the problem states in part b.

yeah I realized it very soon after typing

purebladeknight · #6 $Old$ November 3rd, 2009, 09:47 PM

hi im new but i'll answer your Q to the best of my knowledge & please correct me if i am wrong.

(a) A rectangle has the same area as a square. The long side of the rectangle is four times the length of the short side. The perimeter of the square has length 160 mm. How much longer is the perimeter of the rectangle than the perimeter of the square?

40mm

(b) Same question but this time the perimeter of the square is L mm and the long side of the rectangle is x times as long as the short side.

data: Asq = Ar, Ar = x^3, Asq = Side(sq)^2, Psq = L.

the Perimeter(sq) = Lmm, so Side(s)=Lmm/4 since sq has 4 = sides

now we can use Side(s) = x = Lmm/4 and sub into Asq,

Asq=(Lmm/4)^2 which = Ar = x^3

so, (L/4)^2 = x^3

x = (L/4)^(2/3)

now Pr = 2x + 2x^2

sub in x = (L/4)^(2/3) into Pr = 2x + 2x^2

Pr = 2((L/4)^(2/3)) + 2((L/4)^(2/3))^2

finally Pr - Ps

[2((L/4)^(2/3)) + 2((L/4)^(2/3))^2] - L

Togechu64 · #7 $Old$ November 3rd, 2009, 09:54 PM

I'm not sure exactly where you went wrong but that isn't the correct answer :/

pflo · #8 $Old$ November 3rd, 2009, 10:13 PM

Quote:

Originally Posted by Togechu64 $View Post$

(a) A rectangle has the same area as a square. The long side of the rectangle is four times the length of the short side. The perimeter of the square has length 160 mm. How much longer is the perimeter of the rectangle than the perimeter of the square?
(b) Same question but this time the perimeter of the square is L mm and the long side of the rectangle is x times as long as the short side.

You are looking for the difference in perimeters, or $P_{r} - P_{s}$ .

$P_{r}=2l + 2w$ or $P_{r}=2(xw) + 2w$ . Factoring yields $P_{r}=2w(x+1)$ .

To get the $w$ out of the equation, you can use the fact that areas are the same. The perimeter of the square is $L=4s$ , which means the length of a side is $\frac{L}{4}$ , making the area $\frac{L^2}{16}$ . The area of the rectangle is $(wx)w$ or $w^2x$ . The relationship is then $w^2x=\frac{L^2}{16}$ and $w=\sqrt{\frac{L^2}{16x}}=\frac{L}{4\sqrt{x}}$ .

Plugging this into the formula for the rectangle perimeter, you can see that $P_{r}=2\frac{L}{4\sqrt{x}}(x+1)=\frac{L}{2\sqrt{x}}(x+1)$ .

The differences in perimeters (the answer to the original part b question) is then $\frac{L}{2\sqrt{x}}(x+1)-L$ . You can rationalize this, factor it, distribute, or whatever other simplification you want. The form I kinda like is $L(2\sqrt{x}+\frac{\sqrt{x}}{2x}-1)$ .

Using this final version of the formula to solve part a (plugging in 4 for x and 160 for L), I get 520 mm. If that's the right answer to part a then I'm pretty comfortable with it.

pflo · #9 $Old$ November 3rd, 2009, 10:15 PM

purebladeknight -

I think the error in this is that the area of the rectangle is not x^3. I stopped looking at your solution after that, so if is not the error I'm not sure what it is.

Looking further down it looks like you're saying the length of a side of the square is x (when you say x = L/4). I don't think that's right either.

Togechu64 · #10 $Old$ November 3rd, 2009, 10:21 PM

Quote:

Originally Posted by pflo $View Post$

You are looking for the difference in perimeters, or $P_{r} - P_{s}$ .

$P_{r}=2l + 2w$ or $P_{r}=2(xw) + 2w$ . Factoring yields $P_{r}=2w(x+1)$ .

To get the $w$ out of the equation, you can use the fact that areas are the same. The perimeter of the square is $L=4s$ , which means the length of a side is $\frac{L}{4}$ , making the area $\frac{L^2}{16}$ . The area of the rectangle is $(wx)w$ or $w^2x$ . The relationship is then $w^2x=\frac{L^2}{16}$ and $w=\sqrt{\frac{L^2}{16x}}=\frac{L}{4\sqrt{x}}$ .

Plugging this into the formula for the rectangle perimeter, you can see that $P_{r}=2\frac{L}{4\sqrt{x}}(x+1)=\frac{L}{2\sqrt{x}}(x+1)$ .

The differences in perimeters (the answer to the original part b question) is then $\frac{L}{2\sqrt{x}}(x+1)-L$ . You can rationalize this, factor it, distribute, or whatever other simplification you want. The form I kinda like is $L(2\sqrt{x}+\frac{\sqrt{x}}{2x}-1)$ .

Using this final version of the formula to solve part a (plugging in 4 for x and 160 for L), I get 520 mm. If that's the right answer to part a then I'm pretty comfortable with it.

I'm not sure what the discrepancy is but my friend and I just finished working it out and we got the correct answer:

(2x*(L/4)*sqrt(1/x)+2*L/4*sqrt(1/x))-L

The answer for part (a) was not 520mm however, it was 40mm. Thanks alot for all the help everyone!

pflo · #11 $Old$ November 3rd, 2009, 10:30 PM

That is the same answer I got - before I simplified. My mistake was in the simplification somewhere.