Question 577610
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You are really not doing a very good job of explaining what you want to do, but here is my guess.  I think you want to create a right triangle with two sides that have integer measures such that the third side meassures *[tex \Large \sqrt{15}]


Let *[tex \LARGE a] represent the measure of the short leg of the triangle, let *[tex \LARGE b] represent the measure of the other leg of the triangle, and let *[tex \LARGE c] represent the measure of the hypotenuse.


Pythagoras in its pristine form says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ a^2\ +\ b^2]


But that same relationship can be expressed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ =\ c^2\ -\ a^2]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \sqrt{c^2\ -\ a^2]


What we need to find is two integers such that the difference of their squares is 15.  What if *[tex \LARGE a\ =\ 1]?  Then *[tex \LARGE a^2\ =\ 1], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ -\ 1\ =\ 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 4]


Ah ha!  Two integers.


So we need to construct a right triangle that has sides of *[tex \LARGE 1], *[tex \LARGE \sqrt{15}], and *[tex \LARGE 4] and the right angle has to be between the two shorter segments.


On your graph paper, plot the point (1, 0).  Draw a heavy line from the origin to this point.  Use the scale on your graph paper to set the points of your compass 4 units apart.  Put the metal point of your compass on the point (1,0) then strike a short arc that intersects the positive *[tex \LARGE y]-axis.  This point of intersection will be the point *[tex \LARGE \left(0,\sqrt{15})\right)].


Hints for the other two:  24 = 25 - 1  and 34 = 25 + 9


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