Why is the area of an equilateral triangle $$\frac{\sqrt{3}}{4}b^2$$ and not $$\frac12b^2$$or $$bh$$
2 Answers
$\begingroup$An equilateral triangle has three sides of equal length and three internal angles which are each $60^\circ$.
Imagine cutting the equilateral triangle in half by an altitude. This way, there are two right triangles which have the angle pattern $30^\circ−60^\circ−90^\circ$. This means the sides are in a ratio of $1:\sqrt{3}:2$.
If the altitude is drawn in, the base of the triangle is bisected, leaving two congruent segments with length $1/2$. The side opposite the $60^\circ$ angle, the height of the triangle, is just $\sqrt{3}$ times the existing side of $1/2$, so its length is ${\sqrt{3}}/2$.
Then, since the area of a triangle is $A=\frac{1}{2}bh$, we may deduce
The ratio of side lengths in a $30-60-90$ triangle is $\frac12b:\frac{\sqrt3}2b:b$, where $b$ is the hypotenuse. Its area, therefore, is $$\frac12\cdot\frac12b\cdot\frac{\sqrt3}2b=\frac{\sqrt3}8b^2.$$
Reflect the triangle across its $\frac{\sqrt3}2b$ side, and we get an equilateral triangle. Its area would be double the area above, or $$2\left(\frac{\sqrt3}8b^2\right)=\frac{\sqrt3}4b^2$$as desired.
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