Question 1204421
<pre>
For enrichment, here is a different approach.

{{{drawing(400,2960/9,-5,40,-5,32,

line(0,0,32,0),line(32,0,32,3),line(32,3,0,3),line(0,0,0,3),
line(0,3,0,27),line(0,27,3,27),line(3,27,3,3),line(0,0,35,0),
line(3,27,35,27),line(35,27,35,24),line(35,24,3,24),line(0,0,35,0),

line(32,0,35,0),line(35,0,35,24),line(32,0,32,24),line(0,0,35,0),

locate(15,0,35-x),locate(35.2,13.5,27-x), locate(35.2,26.2,x),
locate(33.5,0,x)


 )}}}

Let the width of the strip be x. 
The area of the floor is 27x35 = 945, and the area of the rug is 609,
So the total area of the strip is 945-609 = 336

As you see from the drawing above, the strip is made up of two
vertical 27-x by x thin rectangles and vertical two horizontal
rectangles. So we have the equation:

{{{2x(27-x)}}}{{{""+""}}}{{{2x(35-x)}}}{{{""=""}}}{{{336}}}

{{{54x-2x^2+70x-2x^2}}}{{{""=""}}}{{{336}}}

{{{-4x^2+124x-336}}}{{{""=""}}}{{{0}}}

Divide through by -4

{{{x^2-31x-84}}}{{{""=""}}}{{{0}}}

That's the same quadratic equation the other tutors got, so the
rest is the same as theirs.  I'll stop here.

Edwin</pre>