Question

Question 6
Find the altitude of an equilateral triangle of side 8cm.

Answer 1

Let ABC be an equilateral triangle of side 8 cm i.e., AB = BC = CA = 8 cm.
Draw altitude AD which is perpendicular to BC. Then, D is the mid – point of BC.

$\therefore BD=CD=\frac{1}{2}BC=\frac{8}{2}=4cm$
$Now,A{B}^2=A{D}^2+B{D}^2\left[byPythagorastheorem\right]$
$\Rightarrow (8{)}^2=A{D}^2+(4{)}^2$
$\Rightarrow 64=A{D}^2+16$
$\Rightarrow A{D}^2=64-16=48$
$\Rightarrow AD=\sqrt{48}=4\sqrt{3}cm$.
Hence, altitude of an equilateral triangle is $4\sqrt{3}cm$.