Question 1126663
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Presuming that you are using the standard and commonly accepted meaning of the term "length", then you have reduced the 8 cm dimension to 2 cm, a factor of one-fourth.  If you reduce the other dimension by the same ratio, then since the area is calculated by multiplying the length times the width the ratio of the areas would be the product of the reduction ratios of the two dimensions.  In other words, one-fourth times one-fourth.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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