Question 1048957
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The length of a rectangle is 8 cm more than its width. If the length is decreased by 9 and the width is tripled, 
the area is increased by 50%. What was the area of the original {{{highlight(cross(triangle))}}} rectangle?
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{{{L=w+8}}}
{{{wL=A}}}, original area


Changed rectangle, {{{(L-9)(3w)=(3/2)A}}}, the 50% area increase.
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{{{3wL-27w=(3/2)A}}}
{{{3A-27w=(3/2)A}}}
{{{3A-(3/2)A=27w}}}
{{{(3/2)A=27w}}}
{{{(3/2)LW=27W}}}  --->  {{{(3/2)L=27}}}  --->  L = {{{2*27)/3}}} = 18 cm.


Then W = L - 8 = 18-8 = 10 cm. Hence, the area was 10*18 = 180 {{{cm^2}}}.


<U>Answer</U>. The dimensions were 10 cm and 18 cm.  The area was 180 {{{cm^2}}}.
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