Question 1192772
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With the percentage of 58%, it is not likely that the width will be a "nice" number.  So we very probably won't get an exact answer without a calculator.<br>
So the only thing to learn from this problem is how to set the problem up for solving.<br>
Let the uniform width of the out-of-bounds area be x; then the dimensions of the court are 84-2x by 50-2x.<br>
The area of the court is to be 58% = 0.58 of the area of the floor:<br>
{{{(84-2x)(50-2x)=0.58(84*50)}}}<br>
Graph the two expressions on a graphing calculator to find that the width of the out-of-bounds area is approximately 7.4 feet.<br>