# constructing wire frames

• Aug 6th 2007, 04:09 PM
Jonboy
constructing wire frames
Hello I just want to verify this answer:

Two square wire frames are to be constructed from a piece of wire 100 inches long. If the area enclosed by one frame is to be one-half the area enclosed by the others, find the dimensions of each frame. (Disregard the thickness of the wire)

So each sides on the square of smaller area is x
Each sides on the be bigger square would have to be $\displaystyle x\sqrt{2}$

Then I would do: $\displaystyle 4x\,+\,4x\sqrt{2}\,=\,100$

$\displaystyle x(4\,+\,4\sqrt{2})\,=\,100$

$\displaystyle x\,=\,25\,+\,25\sqrt{2}$

Is this fundamentally sound answer? Thanks all.
• Aug 6th 2007, 04:18 PM
topsquark
Quote:

Originally Posted by Jonboy
Hello I just want to verify this answer:

Two square wire frames are to be constructed from a piece of wire 100 inches long. If the area enclosed by one frame is to be one-half the area enclosed by the others, find the dimensions of each frame. (Disregard the thickness of the wire)

So each sides on the square of smaller area is x
Each sides on the be bigger square would have to be $\displaystyle x\sqrt{2}$

Then I would do: $\displaystyle 4x\,+\,4x\sqrt{2}\,=\,100$

$\displaystyle x(4\,+\,4\sqrt{2})\,=\,100$

$\displaystyle x\,=\,25\,+\,25\sqrt{2}$

Is this fundamentally sound answer? Thanks all.

The method is correct, your solution is wrong.
$\displaystyle 4x + 4x \sqrt{2}= 100$

$\displaystyle 4x(1 + \sqrt{2}) = 100$

$\displaystyle x = 25 \cdot \frac{1}{1 + \sqrt{2}}$

$\displaystyle x = 25 \cdot \frac{1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}}$

$\displaystyle x = 25 \cdot \frac{1 - \sqrt{2}}{1 - 2}$

$\displaystyle x = 25(\sqrt{2} - 1)$

$\displaystyle x = 25 \sqrt{2} - 25$

-Dan
• Aug 7th 2007, 01:58 AM
Jonboy
Thanks Topsquark, I completely understand now. :)