A coin of radius 1 cm is tossed onto a plane surface that has been tesselated by right

triangles whose sides are 8 cm, 15 cm, and 17 cm long. What is the probability that the
coin lands within one of the triangles?

Well, that's a "tri-cky" question! To find the probability, we need to compare the area of the triangle to the area of the plane surface. Let's dive into the calculations to solve this conundrum, shall we?

The area of a triangle can be calculated using the formula: (base * height) / 2. In this case, the base of the triangle is 15 cm and the height is 8 cm. So, the area of one triangle is (15 * 8) / 2 = 60 square cm.

Now, let's move on to the area of the plane surface. To calculate the area, we need to know the dimensions of the entire surface. Since you haven't provided those dimensions, I'll have to use my imagination. Let's say the plane surface is a large sheet measuring 100 cm by 100 cm, giving us 10,000 square cm.

The probability of the coin landing within one of the triangles is the area of one triangle divided by the total area of the surface: 60 / 10,000 = 0.006, or 0.6%.

So, the coin has only a "tri-ling" chance of landing within one of the triangles in this scenario.

To find the probability that the coin lands within one of the triangles, we need to determine the area of the region where the coin can land and divide it by the total area of the surface.

Let's start by calculating the total area of the surface. Since the surface is made up of right triangles, we can calculate the area of each triangle and then sum them up.

The area of a right triangle can be calculated using the formula: Area = (base * height) / 2.

In this case, the base and height of each triangle are given as 8 cm and 15 cm respectively. So the area of one triangle is (8 * 15) / 2 = 60 cm^2.

Now, we need to determine the number of triangles on the surface. Since the surface is tessellated by right triangles, we can calculate the number of triangles by dividing the total area of the surface by the area of one triangle.

The total area of the surface is given by the formula: Total area = (base * height).

In this case, the base and height of the surface are given as 8 cm and 15 cm respectively. So the total area of the surface is (8 * 15) = 120 cm^2.

Dividing the total area of the surface by the area of one triangle, we get the number of triangles on the surface: Number of triangles = Total area / Area of one triangle = 120 cm^2 / 60 cm^2 = 2 triangles.

Now, let's calculate the area where the coin can land. Since the coin has a radius of 1 cm, its landing area is a circle with a radius of 1 cm. The area of a circle can be calculated using the formula: Area = π * (radius)^2.

In this case, the radius of the circle is 1 cm. So the area where the coin can land is π * (1)^2 = π cm^2.

Therefore, the probability that the coin lands within one of the triangles is the ratio of the area where the coin can land to the total area of the surface: Probability = Area where the coin can land / Total area of the surface = π cm^2 / 120 cm^2 ≈ 0.008.

So, the probability that the coin lands within one of the triangles is approximately 0.008 or 0.8%.

To find the probability that the coin lands within one of the triangles, we need to compare the area of the region where the coin can land within a triangle, to the total area of the entire plane surface.

First, let's find the area of each triangle using Heron's formula:

Given the sides of the triangle are 8 cm, 15 cm, and 17 cm, the semi-perimeter (s) is calculated as:
```
s = (8 + 15 + 17) / 2 = 20
```

Now we can use Heron's formula to find the area of the triangle (A):
```
A = sqrt(s * (s - 8) * (s - 15) * (s - 17))
```

Next, we calculate the total area of the entire plane surface. Since the triangles are tessellated on the plane surface, we can assume it is infinite. Therefore, the total area is infinite.

Now we can calculate the probability that the coin lands within one of the triangles:
```
P = (Area of the region where the coin can land within a triangle) / (Total area of the entire plane surface)
```
Since the total area is infinite, the probability of landing within a triangle becomes zero.

Therefore, the probability that the coin lands within one of the triangles is zero.

well, if it is entirely within, it cannot touch any corner or side.

STart with the area along the sides the center of the coin cannot be in.

The center has to be greater than .5cm from any side. Draw the original triangle. Now mark a line parallel to each side, inside the triangle, .5 from the parallel side. These three lines make a triangle inside, in which the center may reside.

label the inner triangle a, b, c, with the a side parallel to 8, b parallel to 15, and c parallel to 17.

a/8=b/15=c/17 similar triangles.

but a= 8-.5-.5-.5(8/15) If you examine the 17 corner carefully, you can arrive at that with using smaller triangles which are similar.
solve for a: a= 6.86
which means b: b= a*15/8=12.86

area of inner triangle= 1/2 ab=44.1
area of original triangle= 1/2 8*15=60

Pr(coin inside)= areainner/areaouter