Question

Find the length of the altitude of an equilateral tringle with side 6 cm

Answer 1

Answer:

AD=3 ROOT 3 CM

Explanation:

AB=BC=AC=6 CM            (GIVEN)

draw an altitude AD from vertex A perpendicular to BC

THEREFORE BASE IS DIVIDED INTO 2 PARTS OF 3 CM EACH.

In triangle ADB,

AD^2+BD^2=AB^2

AD^2=6^2-3^2

AD^2=27

AD=3 ROOT 3 CM

Answer 2

Answer:

Altitude = $\sqrt{27}$ or 3$\sqrt{3}$

Explanation:

Let the sides of triangles be AB,AC and BC

We are given that AB=AC=BC=6cm

Therefore altitude will divide triangle side into two equal halves BD=CD=3cm,

By applying Pythagoras theorem in  ΔABD we get,

$AB^{2} - BD^{2} = AD^{2}$

$6^{2} - 3^{2} = AD^{2}$

36 - 9 = $AD^{2}$

27 = $AD^{2}$

AD = $\sqrt{27}$

AD = 3$\sqrt{3}$