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• Sep 20th 2009, 12:38 PM
Vesicant
Been having some trouble with this problem, was wondering if anyone would be able to help me out.

Given: Two rectangles have dimensions in meters expressed as radical equations.

If the difference of the lengths of their diagonals is 2, find the dimensions of the rectangles.

Shown below:

http://i38.tinypic.com/htycmo.jpg
• Sep 20th 2009, 12:43 PM
skeeter
Quote:

Originally Posted by Vesicant
Been having some trouble with this problem, was wondering if anyone would be able to help me out.

Given: Two rectangles have dimensions in meters expressed as radical equations.

If the difference of the lengths of their diagonals is 2, find the dimensions of the rectangles.

Shown below:

http://i38.tinypic.com/htycmo.jpg

use Pythagoras to determine the length of each diagonal, then set up an equation ...

short diagonal + 2 = long diagonal

solve for x, and ATQ.
• Sep 20th 2009, 12:46 PM
Vesicant
Quote:

Originally Posted by skeeter
use Pythagoras to determine the length of each diagonal, then set up an equation ...

short diagonal + 2 = long diagonal

solve for x, and ATQ.

ah pythagoras? I was using the diagonal formula(d^2= width^2+height^2) and got D=5x^2+4x+1

Care to tell me what answer I would get when using the pythagoram theorem so I know that I'm doing it right, I assumed I would have to find the diagonal first and then get the pythagorean equation for the two triangles.?
• Sep 20th 2009, 12:59 PM
skeeter
Quote:

Originally Posted by Vesicant
ah pythagoras? I was using the diagonal formula(d^2= width^2+height^2) and got D=5x^2+4x+1

Care to tell me what answer I would get when using the pythagoram theorem so I know that I'm doing it right, I assumed I would have to find the diagonal first and then get the pythagorean equation for the two triangles.?

set it up using Pythagoras and I'll let you know.
• Sep 20th 2009, 01:07 PM
Vesicant
Set up with pythagoras the first would be

sqrt x^2 + (sqrt2x+1)^2=c^2
x^2+ (2x+1)(2x+1)=c^2
X^2 +4x^2+2x+2x+1=c^2
5x^2+4x+1=c^2

The second:

1^2+sqrt x^2=c^2
1+x^2=c^2

Hows that look?
• Sep 20th 2009, 01:12 PM
skeeter
Quote:

Originally Posted by Vesicant
Set up with pythagoras the first would be

sqrt x^2 + (sqrt2x+1)^2=c^2
x^2+ (2x+1)(2x+1)=c^2
X^2 +4x^2+2x+2x+1=c^2
5x^2+4x+1=c^2

The second:

1^2+sqrt x^2=c^2
1+x^2=c^2

Hows that look?

first of all, $(\sqrt{x})^2 = x$ and $(\sqrt{2x+1})^2 = 2x+1$

• Sep 20th 2009, 01:37 PM
Vesicant
Quote:

Originally Posted by skeeter
first of all, $(\sqrt{x})^2 = x$ and $(\sqrt{2x+1})^2 = 2x+1$

sqrt x^2 + (sqrt2x+1)^2=c^2
x+2x+1=c^2
3x+1=c^2
• Sep 20th 2009, 01:56 PM
skeeter
Quote:

Originally Posted by Vesicant
sqrt x^2 + (sqrt2x+1)^2=c^2
x+2x+1=c^2
3x+1=c^2

ok ... now find $c^2$ for the other rectangle.
• Sep 20th 2009, 01:57 PM
Vesicant
Quote:

Originally Posted by skeeter
ok ... now find $c^2$ for the other rectangle.

1+x= $c^2$

Now what?
• Sep 20th 2009, 02:08 PM
skeeter
Quote:

Originally Posted by Vesicant
1+x= $c^2$

Now what?

well, the first diagonal length is $\sqrt{3x+1}$ , and the second is $\sqrt{1+x}$

I believe the relative sizes of your rectangles in the sketch are backwards. the left one should be larger than the right ...

you know the difference in the lengths between the two diagonals is 2.

$\sqrt{x+1} + 2 = \sqrt{3x+1}$

now solve for x and then determine the dimensions of each rectangle.
• Sep 20th 2009, 02:19 PM
Vesicant
Quote:

Originally Posted by skeeter
well, the first diagonal length is $\sqrt{3x+1}$ , and the second is $\sqrt{1+x}$

I believe the relative sizes of your rectangles in the sketch are backwards. the left one should be larger than the right ...

you know the difference in the lengths between the two diagonals is 2.

$\sqrt{x+1} + 2 = \sqrt{3x+1}$

now solve for x and then determine the dimensions of each rectangle.

Alright so X=2

Then I just input x into all of the original expressions I was given and then multiply width by height to get my dimensions?
• Sep 20th 2009, 02:24 PM
skeeter
Quote:

Originally Posted by Vesicant
Alright so X=2

Then I just input x into all of the original expressions I was given and then multiply width by height to get my dimensions?

x does not equal 2

how did you get that?
• Sep 20th 2009, 02:27 PM
Vesicant
Quote:

Originally Posted by skeeter
x does not equal 2

how did you get that?

$\sqrt{x+1} + 2 = \sqrt{3x+1}$
$(\sqrt{x+1} + 2)^2 = (\sqrt{3x+1})^2$
x+1+4=3x+1
4=2x
4/2=2x/2
X=2
• Sep 20th 2009, 02:42 PM
skeeter
Quote:

Originally Posted by Vesicant
$\sqrt{x+1} + 2 = \sqrt{3x+1}$
$\textcolor{red}{(\sqrt{x+1} + 2)^2} = (\sqrt{3x+1})^2$
x+1+4=3x+1
4=2x
4/2=2x/2
X=2

left side ...

$(\sqrt{x+1} + 2)^2$

$(x+1) + 4\sqrt{x+1} + 4$
• Sep 20th 2009, 02:52 PM
Vesicant
Quote:

Originally Posted by skeeter
left side ...

$(\sqrt{x+1} + 2)^2$

$(x+1) + 4\sqrt{x+1} + 4$

ah I see what I did wrong

http://www.mathhelpforum.com/math-he...1ae514fd-1.gif=3x+1

$(x+1)^2 + 4^2(\sqrt{x+1})^2 + 4^2=(3x+1)^2$
$(x+1)(x+1)+16(x+1) + 16=(3x+1)(3x+1)$
$x^2+2x+1+16x+32=9x^2+6x+1$
$=-8x^2+4x-32$
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