document.write( "Question 368170: The perimeter of a rectangle is 32 inches. The area of the rectangle is 48 square inches. Find the dimension of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #262366 by stanbon(75887) $\"\"$ $\"About$

You can put this solution on YOUR website!
The perimeter of a rectangle is 32 inches. The area of the rectangle is 48 square inches. Find the dimension of the rectangle.
\n" ); document.write( "------------------------
\n" ); document.write( "Equations:
\n" ); document.write( "Area: lw = 48
\n" ); document.write( "Per:: 32 = 2(l + w)
\n" ); document.write( "-----------------------
\n" ); document.write( "lw = 48
\n" ); document.write( "16 = l+w
\n" ); document.write( "----
\n" ); document.write( "w = 16-l
\n" ); document.write( "---
\n" ); document.write( "Substitute for \"w\" and solve for \"l\":
\n" ); document.write( "l(16-l) = 48
\n" ); document.write( "16l - l^2 = 48
\n" ); document.write( "l^2 - 16l + 48 = 0
\n" ); document.write( "-------------
\n" ); document.write( "l^2 -4l-12l + 48 = 0
\n" ); document.write( "l(l-4)-12(l-4) = 0
\n" ); document.write( "(l-4)(l-12) = 0
\n" ); document.write( "If length = 4 inches; then width = 12 inches
\n" ); document.write( "If length = 12 inches; then width = 4 inches
\n" ); document.write( "================================================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "============= \n" ); document.write( "

\n" );