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In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.


Answer :

Given, an equilateral triangle say ABC,
Graph 2
Let the sides of the equilateral triangle be of length a, and AE be the altitude of \(\triangle \) ABC.
BE = EC = \(\frac{BC}{2} = \frac{a}{2} \)
In \(\triangle\) ABE, by Pythagoras Theorem, we get
\(\Rightarrow AB^2 = AE^2 + BE^2\)
\(\Rightarrow a^2 = AE^2 + ({{a} \over {2}})^2\)
\(\Rightarrow AE^2 = a^2 - {{a^2} \over {4}}\)
\(\Rightarrow AE^2 = {{3a^2} \over {4}}\)
\(\Rightarrow 4AE^2 = 3a^2\)

4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.

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