Question 254632
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If the ratio of the corresponding sides is 1:4, then if the measure of the length of the smaller rectangle is *[tex \Large l], then the measure of the length of the larger rectangle is *[tex \Large 4l].  Likewise, the measures of the widths must be *[tex \Large w] and *[tex \Large 4w].


The perimeter of the small rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_S\ =\ 2l\ +\ 2w]


and the perimeter of the large rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_L\ =\ 2(4l)\ +\ 2(4w)]


From the last relationship, factor out a 4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_L\ =\ 4\left(2l\ +\ 2w\right)]


Note that the quantity inside the parentheses is identical to the RHS of the first equation, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_L\ =\ 4P_S]


Hence, the ratio of *[tex \LARGE P_S:P_L] is *[tex \LARGE 1:4]


The area of the smaller rectangle is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_S\ =\ lw]


The area of the larger rectangle is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_L\ =\ (4l)(4w)\ =\ 16lw\ =\ 16A_S]


Ratio is *[tex \LARGE 1:16]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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