Question

. In an equilateral triangle with side a, prove that area =
√ 3/4a^2

an?​

Answer 1

Answer:

We know, that the altitude of an equilateral triangle on a base bisects it.

Let the length of side be a.

So, the altitude divides the base into lengths = a/2 and a/2.

Consider, in a ΔABC, altitude drawn from A bisects BC at D perpendicularly.

Now, consider right Δ ABD.

In it, hypotenuse is the side of equilateral triangle, while the base, which is one side is of length a/2. The other side is the Altitude.

As per Pythagoras Theorem,

(a)^2 = (a/2)^2 + (Altitude)^2     [ Here a is the side of eq. triangle, and a/2 is                      

                                                    the base ]

(Altitude)^2 = a^2 - a^2/4 = 3a^2/4

So, altitude = √3a/2

Now, area of triangle = 1/2 * b * h

Here base = base of equilateral triangle, whose length is a

So, area = 1/2 * a * √3a/2 = √3a^2/4. (proved)