Question

In an equilateral triangle of side $3\sqrt{3}$, find the length of the altitude.

Answer 1

REF.Image

Given

side of an equilateral triangle

$ABC=3\sqrt{3}cm$

$AB=BC-AC=3\sqrt{3}cm$

Let AD=h (altitude)

$BD=\frac{1}{2}B$ (Altitude bisect the base)

$BD=\frac{1}{2}.3\sqrt{3}=3\sqrt{3}/2$ cm

$A{B}^2=A{D}^2+B{D}^2$

$(3\sqrt{3}{)}^2=(h{)}^2+(3\sqrt{3}/2{)}^2$

$\Rightarrow 27=h^2+27/4$

$\Rightarrow h^2=27-27/4$

$\Rightarrow h^2=(4.27-27)/4$

$\Rightarrow h^2=108-27/4$

$\Rightarrow h^2=81/7$

$\Rightarrow h=\sqrt{81/4}$

$\Rightarrow h=9/2$

$\Rightarrow h=4.5cm$

Hence, the length of the altitude h is 4.5 cm