Question 220233
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If you construct an altitude of an equilateral triangle, then the altitude forms a right angle with the base.  That means that you have now formed two 30-60-90 triangles, i.e. right triangles where the hypotenuse is one side of the equilateral triangle and the short leg is exactly one-half the measure of the hypotenuse.  Now having a hypotenuse of 4 and a short leg of 2, we can use Pythagoras to calculate the measure of the long leg which is the same as the altitude of the equilateral triangle:


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



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \sqrt{4^2 - 2^2}\ =\ \sqrt{12}\ =\ \sqrt{4\cdot3}\ =\ \sqrt{4}\cdot\sqrt{2}\ =\ 2\cdot\sqrt{3}]


In general, if the measure of the length of a side of an equilateral triangle is  *[tex \Large s], then the measure of an altitude is:  *[tex \Large s\cdot\frac{\sqrt{3}}{2}]



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