Question

ABC is an equilateral triangle of side 4a units. Find the length of its altitude.

Answer 1

Given: Length of each side of the equilateral triangle = 4a units
Construction: Draw a perpendicular from A to BC. Let this meeting point be D.

In $△ABD$ and $△ACD$,
AB = AC
($\because$ All sides of an equilateral triangle are equal)
AD = AD (Common)
$\angle ADB=\angle ADC={90}^{\circ }$
$\therefore$ $△ABD\cong △ACD$
(By RHS congruency)

$\Rightarrow$ BD = DC = 2a units (by cpct)

In $△ABD$,
$A{B}^2=A{D}^2+B{D}^2$ (By pythagoras theorm)
$\Rightarrow$ $A{D}^2=A{B}^2-B{D}^2$
$\Rightarrow$ $A{D}^2=(4a{)}^2-(2a{)}^2$
$\Rightarrow$ $A{D}^2=16a^2-4a^2$
$\Rightarrow$ $A{D}^2=12a^2$
$\Rightarrow$ $AD=2\sqrt{3}a$ units
$\therefore$ Altitude of an equilateral triangle with side 4a will be $2\sqrt{3}a$ units.