Radical Equations Dimensions Problem

**Vesicant** · September 20th 2009, 12:38 PM

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:

**skeeter** · September 20th 2009, 12:43 PM

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:

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

short diagonal + 2 = long diagonal

solve for x, and ATQ.

**Vesicant** · September 20th 2009, 12:46 PM

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.?

**skeeter** · September 20th 2009, 12:59 PM

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.

**Vesicant** · September 20th 2009, 01:07 PM

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?

**skeeter** · September 20th 2009, 01:12 PM

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

make your corrections

**Vesicant** · September 20th 2009, 01:37 PM

Originally Posted by skeeter

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

make your corrections

sqrt x^2 + (sqrt2x+1)^2=c^2
x+2x+1=c^2
3x+1=c^2

**skeeter** · September 20th 2009, 01:56 PM

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.

**Vesicant** · September 20th 2009, 01:57 PM

Originally Posted by skeeter

ok ... now find $c^2$ for the other rectangle.

1+x= $c^2$

Now what?

**skeeter** · September 20th 2009, 02:08 PM

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.

**Vesicant** · September 20th 2009, 02:19 PM

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?