An architect wants to draw a rectangle with a diagonal of 13 inches. The length of the rectangle is to be 2 inches more than twice the width. What dimensions should she make the rectangle?

Question

zepdrix · Answer

|dw:1354490407180:dw|They told us that the length L should be 2 times as long as the width W, which we can write as L=2W. And then it's also 2 inches more than that (ADD 2).
$$L=2W+2$$

zepdrix · Answer

From here we want to apply the Pythagorean Identity.$$\large W^2+L^2=13^2$$Replacing L with its equivalent form in W that we found earlier.$$\large W^2+(2W+2)^2=13^2$$

zepdrix · Answer

So from here, it's a little bit of annoying algebra to solve for W.
The idea is, we want to be able to solve for ONE of the dimensions, and then we'll be able to easily use that number to find the other dimension.

So in this case, we're just trying to find the correct Width.

zepdrix · Answer

Have an idea of what to do next? :D

anonymous · Answer

Well, I was wondering if for the w^2+(2w+2)^2=13^2 is it 4w^2+8w+4+2w?

zepdrix · Answer

$$\large W^2+(2W+2)^2=13^2$$$$\large W^2+(4W^2+4W+4W+4)=13^2$$$$\large 5W^2+8W+4=13^2$$I think you end up with this equation.

anonymous · Answer

Does that factor?

zepdrix · Answer

Hmm, yes I think it does, gimme one sec.

zepdrix · Answer

Mmmm it probably does, but I can't quite seem to make it work lol.
I got nice numbers when throwing it into the quadratic formula though.
So either method should work out very nicely :D

Oh oh I think I just figured out the factors, they're alil confusing though :D

zepdrix · Answer

$$\large5W^2+8W-165=0$$Hmm 165 = 33 * 5
So it will factor something like this,$$\large (w-5)(5w+33)=0$$

zepdrix · Answer

woops those shouldn't have switched to lowercase w's :P my bad