Question 234893
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The area of an equilateral triangle given the measure of one side, *[tex \Large s], is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{s^2\sqrt{3}}{4}].


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{s^2\sqrt{3}}{4}\ = 16\sqrt{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{s^2}{4}\ = 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ 8]


Therefore the perimeter of the triangle, and the length of the wire is 24.


The perimeter of a rectangle is:


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


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(9)\ +\ 2w\ = 24]


So the width is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ 3]


If the length is 9 and the width is 3, the area must be 27.



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