Question

This figure shows how to create a six-pointed star from twelve equilateral triangle tiles: [asy]

size(7cm);
pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(argu*pi/180)); }

for(int i=90;i<450;i+=60) {
pair c=cis(1.2,i);
path p=c-cis(1,i)--c-cis(1,i+120)--c-cis(1,i-120)--cycle;
fill(p,orange+white);
draw(p);
pair c=cis(2.4,i);
path p=c+cis(1,i)--c+cis(1,i+120)--c+cis(1,i-120)--cycle;
fill(p,orange+white);
draw(p);
};

label("$\longrightarrow$",(4,0));
pair x=(8,0);
real s=sqrt(3);
path p=x+cis(s,0)--x+cis(3,30)--x+cis(s,60)--x+cis(3,90)--x+cis(s,120)--x+cis(3,150)--x+cis(s,180)--x+cis(3,210)--x+cis(s,240)--x+cis(3,270)--x+cis(s,300)--x+cis(3,330)--cycle;
fill(p,orange+white);
draw(p);
[/asy] If each of the original tiles has a perimeter of $10$ cm, what is the perimeter of the final star in cm?

Answer 1

40

Explanation:

The star has $12$ sides. Each side is one-third of the perimeter of a triangular tile, or $\frac{10}3$ cm. So the perimeter of the star is

$$12\cdot\frac {10}3 = 4\cdot 10 = \boxed{40\text{ cm}}.$$

Alternatively, consider that the original tiles are composed of $12$ triangles with $3$ sides each, which have $12\cdot 3 = 36$ sides in all. Only $12$ of those $36$ sides make up the perimeter of the star. $12$ is one-third of $36,$ so the perimeter of the star is one-third of the total perimeter of the tiles. The tiles have a total perimeter of $10 \cdot 12=120\text{ cm},$ so the perimeter of the star is $\frac{120}3 = 40$ cm.

Answer 2

40 cm

Explanation:

Each point of the final 6-pointed star has 2/3 of the perimeter of the equilateral triangle. So, the 6 points have a total perimeter of ...

6(2/3)(10 cm) = 40 cm

The perimeter of the final star is 40 cm.