constructing wire frames

**Jonboy** · August 6th 2007, 04:09 PM

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 $x\sqrt{2}$

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

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

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

Is this fundamentally sound answer? Thanks all.

**topsquark** · August 6th 2007, 04:18 PM

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 $x\sqrt{2}$

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

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

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

Is this fundamentally sound answer? Thanks all.

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

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

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

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

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

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

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

-Dan

**Jonboy** · August 7th 2007, 01:58 AM

Thanks Topsquark, I completely understand now.

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