## College Algebra 7th Edition

$x=5$ $in$
Draw vertical and horizontal lines separating your figure into two rectangular regions in addition to square region. The long rectangular region is $14$ by $x$, so its area is $14x$ $in²$, while the short one is $13$ by $x$, so its area was $13x$ $in²$. The square is $x$ by $x$ so its area is $x^2$ $in²$. The combined areas are : $14x + 13x+x^2 = 160$ $\Rightarrow x^2+27x-160=0$ $\Rightarrow (x-5)(x+32)=0$ $\Rightarrow x=5$ or $x=-32$ $\color{red}{rejected}$ $x=5$ $in$