Question

find the area of an equilateral triangle whose side measures 7 units. find the area to the nearest tenth of square unit

Answer 1

$\bf \textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2\sqrt{3}}{4}~~ \begin{cases} s=side's~length\\[-0.5em] \hrulefill\\ s=7 \end{cases}\implies A=\cfrac{7^2\sqrt{3}}{4}\implies A=\cfrac{49\sqrt{3}}{4} \\\\\\ A\approx 21.21762239271874684571\implies A = \stackrel{rounded~up}{21.2}$