Question

in an equiletral triangle with side a, prove that area of triangle =
$\sqrt{3} \div 4a2$


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Answer 1

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 ]

= a2 - a24

h = 3√a2

Step 3: We know that, Area of a triangle = 12 * Base * Height

= 12 * a * 3√a2

= 3√4a2

The area of a equilateral triangle = 3√4a2.