Question 1167554
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Let the measure of the sides of the square be *[tex \Large x]. Since triangles DFC and AED are formed by the sides of the square and must therefore be parallel to the sides of triangle ABC, triangles DFC and AED must be similar to ABC, and therefore their sides are in proportion.


So if CB is half of AB, then DE must be half of AE and CF must be half of DF.  But since all four sides of the square are equal, CF is also half of CB and EB is half of AB.


So we have the area of the square as *[tex \Large x^2], and then the long side of ABC is *[tex \Large 2x\ +\ x\ =\ 3x] and the short side is *[tex \Large x\ +\ \frac{x}{2}\ =\ \frac{3x}{2}], so the area is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{ABC}\ =\ \frac{3x\cdot\frac{3x}{2}}{2}\ =\ \frac{9x^2}{4}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{\frac{9x^2}{4}}\ =\ \frac{4}{9}]

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


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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