Question

Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.

Answer 1

Answer:

It is  proved ,in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.  3AB² = 4AD²

Step-by-step explanation:

Given:

∆ ABC is an equilateral ∆, in which sides are AB = BC = AC = a units and AD ⊥  BC ,

In ∆ADB ,

AB² = AD² + BD²     (by Pythagoras theorem)

a² =  AD² + (a/2)²  

[BD = 1/2BC, since in an equilateral triangle altitude AD ⊥ bisector of BC ]

a² - a²/4 = AD²

AD² = ( 4a² - a²)/4  

AD² =  3a² /4  

3AB²/4 = AD²

[ AB = a]

3AB² = 4AD²

Hence, it is  proved that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.

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Answer 2

$\huge\bold\pink{Solution}$

=>Let ABC be equilateral triangle.

=>Let AD be perpendicular bisector from A on to BC. So BD = CD = 1/2 BC

•ADC is a right angle triangle.

So,

AC² = AD² + DC²

AC² = AD² + (1/2 AC)²           

AD² = 3/4 AC²  

     $\huge\underline\mathfrak{Hence\:Proved}$

4 AD² = 3 AC²