Ok what you want to do here is this. Draw a rectangle with width 17 and height 11. Now circumscribe a rectangle around that rectangle of uniform width. Assign the variable $x$ as being the uniform width on the path between your inner and outer rectangle. You need to draw this.
What you should be looking at now is an 11 by 17 inner rectangle. The outer rectangle will have width length $17+x+x=17+2x$ (don't need length, but make sure your diagram is right). The height length will be $11+x+x=11+2x$. Note the lengths $x$ occur on either end (these are the corners).
Ok you are almost there. There are at least 5 ways to solve this, it just depends upon how you think. The correct diagram is essential here.
One way would be to take the outer rectangular area, subtract the inner rectangular area, set that all equal to 165, and solve for $x$. In this case your equation will be
$$(11+2x)(17+2x)-17 \cdot 11 =165.$$
You could also try an additive process in more than one way. For example, the outer rectangular height is $11+2x$, and the width is $x$, so the left piece including everything on the path from top to bottom will have area $x(11+2x)$. There are two of these, and so your left and right paths have area $2x(11+2x)$. The two remaining unaccounted for pieces have area $17x$ each, and there are two of these, so solve
$$2x(11+2x)+2 \cdot 17 =165.$$
The answer will be the same. You could also try taking the top and bottom path areas, then add the remaining unaccounted for height portions, but I will save that for you to see.
The width of your path should be $\frac{5}{2}$. Note you will get two solutions to your quadratic. You clearly take the positive solution. You will get the same solution by any of the methods.
I do not feel too bad about the plot spoiler here as you have enough work in front of yourself drawing this diagram, setting up the quadratic, and solving it. If you manage to get through this spoiler successfully I think you will have learned your homework lesson.
Note you are asked for the dimensions of the room, not the area of the room. Draw a picture of the scenario. I cannot emphasize how important this is. Now assign the unknown additional room lengths to the variable $x$.
Here is one way: Put the rug in the center (does not really matter), and let $x$ be the additional floor lengths needed on all 4 sides of the rug. Now you have dimensions $16+2x$, and $21+2x$. Now take that corresponding area in terms of $x$, set it equal to two times 336, and solve for $x$. Doing it this way, your new room dimensions will be $16+2x$, by $21+2x$.
Don't respond until you have drawn a picture.
Best Answer
You have a photograph that measures $w$ by $2w$, thus the area of the photograph is $2w^2$. The frame is 4cm wide which adds 4 cm on the left, top, bottom, and right. So the total frame measures $w+8$ by $2w+8$, with an area of $(w+8)(2w+8)$. We know the following equality:
$$(w+8)(2w+8)=3(2w^2)$$ $$2w^2+24w+64=6w^2$$ $$4w^2-24w-64=0$$ $$w^2-6w-16=0$$ $$(w-8)(w+2)=0$$
And the only positive solution is $w=8$ so the answer is c.
The answer is very similar to Martin's answer, however, he forgets to include the 4cm on both sides of the frame: