An infinte plane is tiled with equilateral triangles of side length 9cm. If you drop a coin of radius 1 cm onto the plane, what is the probability the coin lands within one of the triangles?
"within" is taken to mean "not touching a side"
the center of the coin must land inside of a "nested" inner triangle
whose edges are parallel to the 9 cm triangle and .5 cm inside
the ratio of the areas of the inner and outer triangles is the probability
the side length of the inner triangle is ... 9 - √3
the ratio of the areas is the square of the ratio of the side lengths
To determine the probability that the coin lands within one of the triangles, we need to compare their areas.
First, let's find the area of each equilateral triangle. The formula to calculate the area of an equilateral triangle is given by:
Area = (sqrt(3) / 4) * side^2
In this case, the side length of the equilateral triangle is 9 cm. Plugging this value into the formula, we get:
Area = (sqrt(3) / 4) * 9^2
= (sqrt(3) / 4) * 81
≈ 37.18 cm^2
Now, we can find the area of the coin. Since the coin is a circle, its area is given by:
Area = π * radius^2
In this case, the radius of the coin is 1 cm. Plugging this value into the formula, we get:
Area = π * 1^2
= π cm^2
To find the probability, we need to compare the area of the coin to the area of the triangle.
Probability = Area of Coin / Area of Triangle
Probability = (π cm^2) / (37.18 cm^2)
Using a calculator, we can divide the values:
Probability ≈ 0.0845
Therefore, the probability that the coin lands within one of the triangles is approximately 0.0845, or 8.45%.