Question 432737
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In order for the three small rectangles to be congruent, the long side of each of the small rectangles must measure twice the short side.  Let *[tex \Large x] represent the measure of the short side, then the measure of the long side must be *[tex \Large 2x].  Labling the sides leads us to the conclusion that the short side of the large rectangle is *[tex \Large 2x] and the long side is *[tex \Large 3x].


The area of the larger rectangle is then *[tex \Large 2x] times *[tex \Large 3x], or *[tex \Large 6x^2].  But we know this to be equal to 1350 *[tex \Large cm^2].


Solving for *[tex \Large x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ =\ 1350]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 225]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 15]


The perimeter of the large rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\,\cdot\,2x\ +\ 2\,\cdot\,3x\ =\ 10x]


But since we know that *[tex \Large x\ =\ 15], the perimeter of the large rectangle must be 150 cm.


A square with an equal perimeter must have sides that measure 150 divided by 4 which is 37.5 cm.


The area of such a square is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (37.5\text{ cm})^2]


Since your original measurement, namely the area of the large rectangle was given accurate to the nearest square centimeter, you must present your answer to the same precision.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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