Question 202617
Let the original length = L and the original width = W, so that the original area:
{{{A = L*W}}} but the problem says that the "...rectangle is half as wide as it is long.", so this can be expressed as:
{{{W = (1/2)L}}} Substitute this into the first equation and you get:
{{{A = (1/2)L^2}}}
Now, if we subtract 2 cm from the length (L-2) and 2 cm from the width (W-2) the area is decreased by 68 sq.cm. (A-68). Lets find the new area:
{{{A-68 = (L-2)(W-2)}}} but substitute {{{W = (1/2)L}}} and {{{A = (1/2)L^2}}}
{{{(1/2)L^2-68 = (L-2)((1/2)L-2)}}} Simplifying:
{{{(1/2)L^2-68 = (1/2)L^2-3L+4}}} Subtract {{{1/2L^2}}} from both sides.
{{{-68 = -3L+4}}} Subtract 4 from both sides.
{{{-72 = -3L}}} Divide both sides by -3.
{{{L = 24}}} and
{{{W = (1/2)L}}}
{{{W = 12}}}
The original length of the rectangle is 24 cm.
Check:
Original area:
{{{A[o] = L*W}}} Substitute L=24 and W=12
{{{A[o] = 24*12}}}
{{{A[o] = 288}}}sq.cm.
Now the new area is:
{{{A[n] = (L-2)(W-2)}}} Substitute L=24 and W=12
{{{A[n] = (24-2)(12-2)}}}
{{{A[n] = (22)(10)}}}
{{{A[n] = 220}}}sq.cm.
The difference is:
{{{A[n]-A[o] = 288-220}}}={{{168}}}sq.cm.