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ABC is an equilateral triangle of side 2a. Find each of its altitudes.


Answer :

Given, ABC is an equilateral triangle of side 2a.

figure

Draw, AD \(\perp\) BC

In \(\triangle\) ADB and \(\triangle\) ADC,
AB = AC
AD = AD
\(\angle\) ADB = \(\angle\) ADC [Both are 90°]

Therefore, \(\triangle\) ADB \( {\displaystyle \cong }\) \(\triangle\) ADC by RHS congruence.
Hence, BD = DC [by CPCT]

In right angled \(\triangle\) ADB,
\(AB^2 = AD^2 + BD^2\)
\((2a)^2 = AD^2 + a^2\)
\(AD^2 = 4a^2 – a^2\)
\( AD^2 = 3a^2\)
\(AD = \sqrt{3} a\)

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