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Q16   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.

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Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, BE=CE=\frac{a}{2}

In \triangleAEB, by Pythagoras theorem

AB^2=AE^2+BE^2

a^2=AE^2+(\frac{a}{2})^2

\Rightarrow a^2-(\frac{a^2}{4})=AE^2

\Rightarrow (\frac{3a^2}{4})=AE^2

\Rightarrow 3a^2=4AE^2

\Rightarrow 4.(altitude)^2=3.(side)^2

Posted by

seema garhwal

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