Question 1038448
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*[illustration Square_In_EquiTriangle_(2).jpg]


Since triangle ABC is equilateral, it is also equiangular, hence angle BAC, ABC, and ACB all measure 60 degrees.


Since the inscribed figure is a square, segment EF is parallel to segment DG which is contained in segment AC.  Hence the measure of angles BEF and BFE are also 60 degrees.


Since all three angles of triangle BEF have equal measure, triangle BEF is equiangular and is also equilateral.  Therefore, the measure of segment BE is equal to the measure of segment EF which is given as 1.


Since the inscribed figure is a square, angle EDG is right.  Since angles ADE and EDG form a straight angle, angle ADE must be right also.  Therefore, triangle ADE is a 30-60-90 right triangle.


In a 30-60-90 right triangle, the measure of the short leg is exactly one-half of the measure of the hypotenuse, hence, if we represent the measure of segment AE with the variable *[tex \Large x], then the measure of segment AD must be *[tex \Large \frac{x}{2}].


Referring to our old friend Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x^2\ -\ \left(\frac{x}{2}\right)^2}\ =\ 1]


Solve for *[tex \Large x] and then calculate *[tex \Large x\ +\ 1] to find the measure of segment AB.


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 \ \ \ \ \ \ \ \ \ \  

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