Question 1089694
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Start by drawing out an equilateral triangle ABC. Each point is a vertex of the triangle.
Then plot another point D such that it is the midpoint of one of the sides. 
I made point D the midpoint of segment AB.
In this case, the altitude is the segment from point C to point D.
Let's call the altitude h for now (h for height) 
This is what the drawing should look like


<img src = "https://i.imgur.com/tjhjPyM.png">


Note how D splits AB into two equal halves. 


We have these properties:


AD = DB
AD = 21.5, DB = 21.5 (43/2 = 21.5)
AD+DB = 21.5+21.5 = 43 = AB


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Using that drawing above as reference, and focusing on triangle CDB, let's find what h must equal. 


Because triangle CDB is a right triangle, we can use the pythagorean theorem


a = 21.5
b = h
c = 43


{{{a^2 + b^2 = c^2}}} Start with the pythagorean theorem


{{{(21.5)^2 + h^2 = (43)^2}}} Plug in the given values


{{{462.25 + h^2 = 1849}}} Square the terms


{{{462.25 + h^2 - 462.25 = 1849 - 462.25}}} Subtract 462.25 from both sides 


{{{h^2 = 1386.75}}} Simplify


{{{sqrt(h^2) = sqrt(1386.75)}}} Take the square root of both sides


{{{h = 37.2390923627309}}} Compute the square root (use a calculator)


{{{h = 37.2}}} Round to the nearest tenth


<font color=red>The altitude is roughly 37.2 cm</font>


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