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 altitudes.

Answer 1

Proof the given statement

Let $∆ABC$ is an equilateral triangle with altitude $AE$ , each side of an equilateral $∆=a$

To prove: In an equilateral triangle, three times the square of one side is equal to four times the square of one of its altitudes or $3A{B}^2=4A{E}^2$

So, $BE=\frac{1}{2}a$ [$\because$perpendicular bisects the side in on an equilateral $∆$]

In right $∆AEB$, by Pythagoras theorem

$$\begin{gathered} \Rightarrow {\left(AB\right)}^2={\left(AE\right)}^2+{\left(BE\right)}^2 \\ \Rightarrow {\left(a\right)}^2={\left(AE\right)}^2+{\left(\frac{1}{2a}\right)}^2\left[\because AB=a\mathrm{and}BE=\frac{1}{2a}\right] \\ \Rightarrow {\left(a\right)}^2=AE{}^2+{\left(\frac{a}{2}\right)}^2 \\ \Rightarrow {\left(a\right)}^2–{\left(\frac{a}{2}\right)}^2=AE{}^2 \\ \Rightarrow {\left(\frac{3a}{2}\right)}^2=A{E}^2 \\ \Rightarrow {\left(3a\right)}^2=4A{E}^2 \\ \Rightarrow 3A{B}^2=4A{E}^2\left[\because AB=a\right] \end{gathered}$$

Hence, In an equilateral triangle, three times the square of one side is equal to four times the square of one of its altitudes.

Hence Proved.