Finding the Minimum Number of Sides for a Rotating Regular Polygon

[icystrike]

Homework Statement

A floor tile has the shape of a regular polygon. If the tile is removed from the floor
and rotated through 50◦ it will fit back exactly into its original place in the floor.
The least number of sides that the polygon can have is?

I don't know what are the theories that i should be learning to solve this question.

One of the Roman dice in the British Museum has 6 square faces and 8 triangular
faces. It is twice as likely to land on any given square face as any given triangular
face. What is the probability that the face it lands on is triangular, when thrown?

I don't understand what the question requires

Homework Equations

The Attempt at a Solution

 

[Mentallic]

For the first one, if the polygon is rotated by 50^o and is symetrically the same to fit back into the tile space, it means that the least possible sides must be $$\frac{360}{50}=7\frac{1}{5}$$ sides. But it has to be a polygon, thus the answer needs to be an integer. What highest common factor of 50 is also a factor of 360? What does this tell you?

For the second, if a square face is twice as likely as a triangle face, can't you then consider 1 square to = 2 triangles?

 

[icystrike]

For the first one, if the polygon is rotated by 50^o and is symetrically the same to fit back into the tile space, it means that the least possible sides must be $$\frac{360}{50}=7\frac{1}{5}$$ sides. But it has to be a polygon, thus the answer needs to be an integer. What highest common factor of 50 is also a factor of 360? What does this tell you?

For the second, if a square face is twice as likely as a triangle face, can't you then consider 1 square to = 2 triangles?

Thanks!
1)HCF of 360 and 50 is 1800
Therefore the sides of the polygon is 1800/50=36
2)The probability is 8/(6*2+8)=2/5


Can you please prompt me on this question?
If f(x)=$$a_0+a_1x+a_2x^2 ... a_nx^n$$
f(1)=8
f(35)=6^6

Find f(7)

 

[Mentallic]

well you actually found the lowest common denominator of both numbers, but you adjusted your last part to the problem to get the correct answer
Another way, the highest common factor would be 10^o (10 is the largest number that can be multiplied by an integer to become both 50 and 360) and thus the sides of the polygon is 360^o/10^o=36 sides.

Yes you correctly found the probability.

Sorry I'm unsure of the answer myself. You should create another thread for it.

 

[icystrike]

well you actually found the lowest common denominator of both numbers, but you adjusted your last part to the problem to get the correct answer
Another way, the highest common factor would be 10^o (10 is the largest number that can be multiplied by an integer to become both 50 and 360) and thus the sides of the polygon is 360^o/10^o=36 sides.

Yes you correctly found the probability.

Sorry I'm unsure of the answer myself. You should create another thread for it.

You've been a great help , Thanks!