BuilderBoi

\(\sqrt2 \approx 1.41421\)

\(20 \div 1.41421 \approx 14.1421\)

Thus, the smallest integer that works is \(\color{brown}\boxed{15}\)

The same logic leads to the right answer, but Kakashi read the problem wrong. They taught it was within 1 cm, not 2. 

\(\sqrt {25} = 5\)

\(5 \times 2 = 10\)

A works, and E also does, because \(\sqrt{100} = 10\)

p = cost of pie

c = cost of cake

2p = c     (1)

2c + 3p = 56      (2)

Substitute c for 2p: 

4p + 3p = 56

Solve for p, and subsitute it with the first equation. 

\(\color{brown}\boxed {y={1 \over4}}\)

Oh wait, never mind, I think I know how to solve it. 

The circular segment is equal to the area of the region \(ABO-\triangle{ABO}\). 

We know the circular region's area is \({30 \over360} \times 36\pi =3\pi\). 

\(\triangle{ABO}\) is isosceles, with degrees of 30, 70, 70, and 2 sides with 6. Using trig, we find that the area of the triangle is 9. 

Thus, the area of the circular region is: \(\color{brown}\boxed{3\pi - 9}\)

The triangle side length rule states that the 2 shorter side lengths must always be greater than the biggest side. 

Here are all the combinations: 

6-5-4  5-4-3   4-3-2

6-5-3  5-4-2   

6-5-2

6-4-3

Thus, there are only \(\color{brown}\boxed{7}\) ways to form a triangle.