Prove that the altitude of an equilateral triangle of side 2a is \(\sqrt{3} a\).

Solution :

Given : \(\triangle\) ABC, in which each side is of length 2a.triangles

To Find : AD (altitude)

In \(\triangle\) ADB and \(\triangle\) ADC,

AD = AD            (common)

\(\angle\) 1 = \(\angle\) 2               (90 each)

AB = AC               (given)


\(\triangle\) ADB \(\cong\) \(\triangle\)  ADC

So,  BD = DC             (By C.P.C.T)

\(\implies\) BD = DC = a

Now, in \(\triangle\) ADB,

\({AD}^2\) + \({BD}^2\) = \({AB}^2\)

\(\implies\)  \({AD}^2\) + \(a^2\) = \((2a)^2\)            (By Pythagoras Theorem)

\({AD}^2\) = \(3a^2\)   \(\implies\)   AD = \(\sqrt{3} a\)

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