Question

5.
Find the length of the altitude of an equilateral
triangle with side 6 cm. [Mar 17, 18​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\:\:$

$\green{\underline \bold{Given :}}$

$\:\:$

• Length of side = 6 cm

$\:\:$

$\red{\underline \bold{To \: Find:}}$

$\:\:$

• length of altitude.

$\rule{200}2$

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

$\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(-3,-3){\line(1,0){39.5}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(17,-3){\line(0,1){20}}\put(18,5){altitude }\put(3,5){}\put(13,-5){base = 6cm}\end{picture}$

$\:\:$

In an equilateral triangle ABC, the altitude AD on side BC is also the median.

So, BD = DC

$\underline{\bold{\texttt{Now, in right triangle ADC}}}$

$\:\:$

$\sf \implies AC^2 = AD^2 + DC^2$

$\:\:$

$\sf \implies AC^2 = AD^2 + \frac { AC^2} { 4 } [Since, AC = BC = 2DC]$

$\:\:$

$\sf \implies AD^2 = AC^2 - \frac { AC^2} { 4}$

$\:\:$

$\sf \implies AD^2 = \frac { 3(AC^2)} { 4}$

$\:\:$

$\sf \implies AD = \frac { (\sqrt3)AC } { 2}$

$\:\:$

Now, we know the value of AB = AC = BC = 6cm

$\:\:$

$\rm So, AD = (\sqrt3) × \frac { 6} { 2} cm$

$\:\:$

$\sf \implies (3\sqrt3) cm$

Hence altitude is $\purple{3\sqrt3 cm}$

$\rule{200}4$