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

1 answer:

8 0

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

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Answer:

Step-by-step explanation: