Proof for area of an equilateral triangle with respect to one side?

I'm trying to find out the area of an equilateral triangle with respect to one side. Anything wrong with my proof?

An equilateral triangle with sides of length $$a$$ can be divided in half along the base to create two triangles with right angles. The sides would be $$h$$ (height), $$y$$ (hypotenuse) and $$b$$ (base). I know what some of these sides are:

$$y = a \\ b = 0.5a \\ h = ?$$

Using Pythagorean theorem, I can find the length of $$h$$:

\begin{align}c^2 &= a^2 + b^2 \\ h^2 &= b^2 + y^2 \\ &= 0.5a^2 + a^2 \\ h &= \pm\frac{a\times\sqrt{3}}{2} \end{align}

Since the length can only be positive, we're left with $$h =\frac{a\times\sqrt{3}}{2}$$.

The area $$A$$ of a triangle is $$A = (b \times h) / 2$$. We can plug in our numbers:

\begin{align} A &= \frac{b \times h}{2} \\ &= \frac{\frac{a}{2} \times \frac{a\times\sqrt{3}}{2}}{2} \\ &= \frac{a^2\times\sqrt{3}}{8} \end{align}

But since we halved the equilateral triangle together, the area of the equilateral is twice this number, since we recombine the two triangles:

\begin{align} A &= 2\times\frac{a^2\times\sqrt{3}}{8} \\ &= \frac{a^2\times\sqrt{3}}{4} \end{align}

This is indeed the area of an equilateral triangle with respect to side $$a$$.

• Heron’s formula for the area of a triangle given its side lengths leads to a particularly simple proof. – amd Jan 21 at 21:26

Overall, this seems fine. Some notes:

• Depending on the intent behind this, you might need to specify that the height you constructed is the median of an isosceles triangle (since under the typical definition, equilateral triangles are isosceles). This thought is mostly drawn from how some profs can be with homework assignments and assuming stuff, I doubt anyone with a decent knowledge of basic geometry would actually doubt it.

• You could probably shave a few bits of calculation/explanation by noting, once you have $$h = a \sqrt 3 / 2$$, you have the base $$b = a$$ as given, and thus $$A = bh/2 = a^2 \sqrt 3 / 4$$ (thus circumventing the whole "this triangle is the area of two half-triangles" bit). Not that your approach is wrong, though, just a little longer in that respect than is necessary.

• There's no real need to introduce $$y$$ as a separate variable, since $$y = a$$ by construction. So you can just note that the hypotenuse is the side of the triangle, and thus $$a$$. Minor nitpick, nothing big.

Overall, just minor nitpicks. Those aside your proof seems good.

Your proof is fine, but in the spirit of visual geometry, let me present a cute alternative: (You might want to break out paper/pencil to follow along with the diagram, but maybe not since the figure is so simple.)

Draw a regular hexagon made from 6 equilateral triangles, then draw the equilateral triangle formed by taking every other vertex of the hexagon. Now, since all equilateral triangles are similar, the area of the large equilateral triangle is $$k^{2}A$$, where $$A$$ is the area of the smaller equilateral triangles, for some scale factor $$k$$. On the other hand, the large equilateral triangle is half the area of the hexagon. So, since the hexagon has area $$6A$$, we have that $$k^{2} = 3$$. Now, we can compute the area of the hexagon as 3 times the area of the rhombuses formed by adjacent pairs of smaller equilateral triangles. The area of each rhombus is $$\frac{ka \cdot a}{2}$$, if the side length of each small equilateral triangle is $$a$$. Thus, we have $$\frac{1}{2}ka^{2} = 6 A$$ and $$k = \sqrt{3}$$, so $$A = \frac{\sqrt{3}}{4}a^{2}$$.