## College Physics (4th Edition)

Let the original area $A_0$ of the poster be $A_0 = L_0\times W_0$ If the length and width are reduce by 20.0%, then the new length and width are $0.80~L_0$ and $0.80~W_0$. We can find the ratio of the new area $A$ to the original area $A_0$: $ratio = \frac{A}{A_0} = \frac{(0.80~L_0)(0.80~W_0)}{(L_0)(W_0)} = (0.80)(0.80) = 0.64$ The new area of the poster is only 64% of the original area which is a decrease of 36% (since 100% - 64% = 36%). The area of the poster must be reduced by a percentage of 36%.