Question

in an equilateral triangle prove that three times the square of one side is equal to four times the square of one of its altitude

Answer 1

Let a is the length of the side of equilateral triangle and AE is the altitude.

So BE = EC = BC/2 = a/2

Now in triangle ABC,

From Pythagoras Theorem

      AB2 = AE2 + BE2

=> a2 = AE2 + (a/2)2

=> AE2 = a2 - a2 /4

=> AE2 = 3a2 /4

=> 4AE2 = 3a2

=> 4*(square of altitude) = 3*(square of one side)

So three times the square of one side is equal to four times the square of one of its altitudes.

Answer 2

given,

let ABC be the equilateral triangle with side 'a' and let AB be it's altitude

to prove,

3 x square of one side =4 x square of it's

altitude

~3a ^2=4AD^2

proof,BD=DC (perpendicular of an equilateral triangle bisect the opposite side)

BD=DC=1/2 BC

BD=DC=a/2

in ADB,

by Pythagoras theorem,

AB^2=AB^2+BD^2

a^2= AD^2+a^2/4

4a^2-a^2/4=AD^2

3a^2=4AD^2

hence proved....