Question 633638
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Construct a square of side measure *[tex \LARGE a] in the first quadrant of the *[tex \LARGE \mathbb{R}^2] plane such that one vertex of the square is at the origin and the two sides adjacent to that vertex are cooincident with the *[tex \LARGE x] and *[tex \LARGE y] axes.


By definition of a square, the four vertices of the square will be *[tex \LARGE A(0,0)], *[tex \LARGE B(a, 0)], *[tex \LARGE C(a,a)], and *[tex \LARGE D(0,a)]


Let point E be represented by the ordered pair *[tex \LARGE (x,y)]


The altitude of triangle ABE to side AB is simply the *[tex \LARGE y] coordinate of point E, namely *[tex \LARGE y].  The measure of the base of triangle ABE is the distance from *[tex \LARGE (0,0)] to *[tex \LARGE (a,0)], which is simply *[tex \LARGE a].  The area of a triangle is given by one-half the altitude times the base, or in the case of triangle ABE, *[tex \LARGE \frac{ay}{2}]


Likewise it can be shown that the area of triangle CDE is given by *[tex \LARGE \frac{a(a\ -\ y)}{2}\ =\ \frac{a^2\ -\ ay}{2}]


The sum of these two areas is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{ay}{2}\ +\ \frac{a^2\ -\ ay}{2}\ =\ \frac{ay\ +\ a^2\ -\ ay}{2}\ =\ \frac{a^2}{2}]


But the area of the square is *[tex \LARGE a^2], so the sum of the areas of the two shaded triangles is in fact one-half of the area of the square.


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