What is the area of an equilateral triangle that has a side length of 4?

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Answer 1 · 2015-12-03T02:23:40

$A = 6.93$ $A= 6.93$ or $4 \sqrt{3}$ $4sqrt3$

$A = \frac{\sqrt{3}}{4} a^{2}$ $A=sqrt3/4a^2$

a $\to$ $rarr$ side which $4$ $4$

$A = \frac{\sqrt{3}}{4} 4^{2}$ $A=sqrt3/(4)4^2$

$A = \frac{\sqrt{3}}{4} 16$ $A=sqrt3/(4)16$

$A = \frac{16 \sqrt{3}}{4}$ $A=(16sqrt3)/4$

$A=(cancel4(4)sqrt3)/cancel4$

$A=4sqrt3$

$sqrt3$ $rarr$ $1.73205080757$

$4sqrt3 = 6.92820323028$

$A= 6.93$