What the area of an equilateral triangle if the side length is 6 mm?

Question

Jayden Jackson · Accepted Answer

#9sqrt3# #"mm"^2#

We can see that if we split an equilateral triangle in half, we are left with two congruent equilateral triangles. Thus, one of the legs of the triangle is #1/2s#, and the hypotenuse is #s#. We can use the Pythagorean Theorem or the properties of #30˚-60˚-90˚# triangles to determine that the height of the triangle is #sqrt3/2s#.

If we want to determine the area of the entire triangle, we know that #A=1/2bh#. We also know that the base is #s# and the height is #sqrt3/2s#, so we can plug those in to the area equation to see the following for an equilateral triangle:

#A=1/2bh=>1/2(s)(sqrt3/2s)=(s^2sqrt3)/4#

In your case, the area of the triangle is #(6^2sqrt3)/4=(36sqrt3)/4=9sqrt3# #"mm"^2#.

Maverick Smith · Answer

undefined The area of an equilateral triangle with side length $ s $ is given by the formula:

$$ \text{Area} = \frac{\sqrt{3}}{4} \times s^2 $$

Substituting the given side length $ s = 6 $ mm into the formula:

$$ \text{Area} = \frac{\sqrt{3}}{4} \times (6)^2 $$

$$ \text{Area} = \frac{\sqrt{3}}{4} \times 36 $$

$$ \text{Area} = \frac{\sqrt{3} \times 36}{4} $$

$$ \text{Area} = \frac{36\sqrt{3}}{4} $$

$$ \text{Area} = 9\sqrt{3} $$

So, the area of the equilateral triangle is $ 9\sqrt{3} $ square millimeters.