Question

Question 6
ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer 1

Given ABC is an equilateral triangle of side 2a.
$DrawAD\perp BC$.

In $\Delta ADB$ and $\Delta ADC$, we have
AB = AC [Given]
AD = AD [Common]
$\angle ADB=\angle ADC$ [each angle equal to ${90}^{\circ }$]
Therefore, $\Delta ADB\cong \Delta ADC$ by RHS congruency rule.
Hence, BD = DC [by CPCT]

In right angled $\Delta ADB$,
$A{B}^2=A{D}^2+B{D}^2$
$\Rightarrow (2a{)}^2=A{D}^2+a^2$
$\Rightarrow A{D}^2=4a^2-a^2$
$\Rightarrow A{D}^2=3a^2$
$\Rightarrow AD=\sqrt{3}a$