How do I find the length of sides and the height of an equilateral triangle when I only know the three angles and the area.
The area is 50.3144 and obviously all the angles are 60 degrees.
I'm in grade 10 so I don't want some crazy formula to figure this out, I've looked everywhere on the internet but all websites talk about SAS, AAS etc. but I have not been given any side length just the area and angles.
[Math] How to find the sides of an equilateral triangle given all angles.
geometry
Related Solutions
You seem to suggest that you want a pentahedron whose five faces are all triangular. There is no such pentahedron, not with equilateral triangles and not with any other kind of triangles.
Every polyhedron whose faces are all triangular has an even number of sides. For example, the regular tetrahedron (4 sides), the octahedron (8 sides), the snub disphenoid (12 sides), and the rest of these.
This is easy to see. Suppose there are $F$ faces and $E$ edges. If we count the 3 edges on each face, we count $3F$ edges total. But this counts each edge twice, because each edge belongs to 2 faces. So the correct number of edges is $E = \frac12\cdot 3F$. But $E$ must be an integer, so $F$ must be even.
In particular, $F=5$ gives $E=\frac{15}2$, which is impossible.
Let us divide ⊿ABC into the following color coded diagram.
For simplicity, we further let [yellow] = 1h, [pink] = 1k, [⊿ABC] = 630 and [green] = x
Then, [orange] = 2h and [red] = 2k.
1 : 2 = [⊿ABG] : [⊿ACG] = (1h + 2h + 1k) : (x + 2k)
After simplification, x = 6h.
∴ [yellow] $= \frac {1}{1 + 6} \cdot \frac {1}{1 + 2} \cdot 630 = 30$
The areas of the other parts can then be found and the results are shown in fig. 2.
In fig.3, it is not difficult to see that [light green] = [green] – [yellow] = … = 150
Also, [⊿HIJ] = [blue] = [red] – [light green] = … = 90, showing that [⊿HIJ] : [⊿ABC] = 1: 7
Best Answer
Let the triangle be $\;\Delta ABC\;$, say with $\;A\;$ the upper vertex, and let $\;D\;$ be the midpoint of side $\;BC\;$ , say. Then, $\;\Delta ABD\;$ is a straight $\;30-60-90\;$ triangle , so $\;|BD|=\frac12|AB|\;$ and $\;|AD|=\frac{\sqrt3}2|AB|\;$ (use Pythagoras Theorem) , and thus the whole triangle's area is
$$\frac{|BC|\cdot |AD|}2=\frac{|AB|\cdot\frac{\sqrt3}2|AB|}2=\frac{\sqrt3}4|AB|^2\stackrel{\text{given}}=50.3144$$
So now just get (fill in details)
$$|AB|=\sqrt{\frac{4\cdot50.3144}{\sqrt3}}$$