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 A_{rectangle} = l·w

If 1/6 of the length and width were used for the miniature, we need to multiply both the length and width by 1/6.

 A_{rectangle} =  \frac{1}{6}  l ·  \frac{1}{6}  w

 A_{rectangle}  \frac{1}{36} lw

So, the area of the miniature was reduced to  \frac{1}{36} of the original area.

720· \frac{1}{36} = 20

Therefore, the area of the miniature is 20 in²