Question

Question 16
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer 1

Let the side of an equilateral triangle be a, and AE be the altitude of $\Delta ABC$.
$\therefore BE=EC=\frac{BC}{2}=\frac{a}{2}$
Applying Pythagoras theorem in $\Delta ABE$, we get
$A{B}^2=A{E}^2+B{E}^2$
$\Rightarrow$ $a^2=A{E}^2+{\left(\frac{a}{2}\right)}^2$
$\Rightarrow$ $A{E}^2=a^2-\frac{a^2}{4}$
$\Rightarrow$ $A{E}^2=\frac{3a^2}{4}$
$\Rightarrow$ $4A{E}^2=3a^2$
Where AE is the altitude and 'a' is the side of an equilateral triangle.
Therefore, $\Rightarrow 4\times$ (Square of altitude) = $3\times$ (Square of one side)