document.write( "Question 517102: The bottom of Jim's rectangular bait box is 3 inches longer than it is wide. The diagonal is 15 inches. What is the area of the bottom of the box??? \n" ); document.write( "
Algebra.Com's Answer #344675 by htmentor(1343)\"\" \"About 
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The bottom of Jim's rectangular bait box is 3 inches longer than it is wide. The diagonal is 15 inches. What is the area of the bottom of the box???
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\n" ); document.write( "Let w = the width of the box
\n" ); document.write( "Then the length = w + 3
\n" ); document.write( "Since the diagonal is 15, we have a right triangle with sides w, w+3 and hypotenuse 15
\n" ); document.write( "Solve using the Pythagorean theorem:
\n" ); document.write( "w^2 + (w+3)^2 = 15^2
\n" ); document.write( "This simplifies to
\n" ); document.write( "w^2 + 3w - 108 = 0
\n" ); document.write( "which can be factored as:
\n" ); document.write( "(w+12)(w-9) = 0
\n" ); document.write( "Take the positive solution: w = 9
\n" ); document.write( "So the area = 9*12 = 108 sq. in.\r
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