Question

A side of an equilateral triangle ABC is $4$ cm . Find the side of an equilateral triangle PQR whose area is double the area of $△ABC$

Answer 1

Area of an equilateral triangle of side-length $a$ is equal to: $\frac{\sqrt{3}}{4}{a}^2$
Area of $\Delta ABC$=$\frac{\sqrt{3}}{4}{4}^2=4\sqrt{3}$
Let $s$ be the side-length of the equilateral triangle with double the area, then:
$\frac{\sqrt{3}}{4}{s}^2=2\times 4\sqrt{3}=8\sqrt{3}$
$⟹s^2=32⟹s=\sqrt{32}=\sqrt{16\times 2}$
$=4\sqrt{2}$