Question

In an equilateral triangle with side a, prove that area =(a root 3)/4

Answer 1

Let, ABC be an equilateral Δ, whose each side measures a units. Now draw AD ⊥ BC.

∴ The altitude and median coincide in an equilateral Δ

∴ BD = DC

⇒ BD = a/2

In right angled Δ ABD, by Pythagoras theorem,

AB² = BD² + AD²

⇒ AD² = AB² - BD²

           = (a)²- (a/2)²

           = a² - (a²/4)

           = (4a²-a²)/2

           = 3a²/2

∴ AD = 3a²/2

Hence,  altitude = 3a²/2

∴ Area of ΔABC = (1/2)× Base × Height

                           = (1/2)× a × (√3a/2)

                           = √3a²/4

Hence, area of ΔABC is = √3a²/4 (is proved)

Answer 2

Step-by-step explanation:

Step 1: Since all the 3 sides of the triangle are same,

AB = BC = CA = a

Step 2: Find the altitude of the △ABC.  

Draw a perpendicular from point A to base BC, AD ⊥ BC

By using Pythagoras theorem

In △ ADC

h2 = AC2 - DC2

= a2 - (a2)2 [Because, DC = a2