If you have a regular pentagon with all equal sides and you split that pentagon into 5 triangles (inside of that pentagon) would the triangles be equalateral or not?
Pentagons and Triangles
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On
In general, any regular polygon, when divided into similar triangles (assuming you use the center as the third point), looks like (apologies for sloppy sketch):
And you know:
- The sum of all the angles in the center is 360 degrees.
- All the angles in the center are equal (because all the triangles are similar).
- The sum of all the angles in a triangle is 180 degrees, and all three angles in an equilateral triangle are the same, therefore an equilateral triangle must have $180/3=60$ degree angles.
So:
- You can compute the center angle of each triangle as $360 / n$ where $n$ is the number of sides.
And this is all you really need to compute. You can find the outer two angles of the triangle if you want, but since you know they are the same, it's sufficient to show whether or not the center angle is 60.
So all you need to do for your problem, or any like it, is compute that center angle, and if $360 / n = 60$, then the triangles are equilateral, otherwise they are not. I'll leave the computations for the 5-sided pentagon up to you.
On
We can draw a picture:
The red angle is the internal angle of a pentagon, and the green angle is half of this, $54^\circ$, which is not $60^\circ$ which would make it an equilateral triangle
We can also see that there are $5$ triangles meeting in the centre. So we calculate $360/5=72^\circ$ also meaning that the triangles are not equilateral


Hint: $$\frac{360}{5}=72\neq 60$$