Right triangle XYZ has legs of length XY = 12 and YZ = 6. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 20?

Guest May 12, 2023

#1**0 **

To solve this problem, we first need to find the area of triangle XYZ. Since triangle XYZ is a right triangle, we can use the formula for the area of a triangle:

Area = (1/2) * base * height

The base of triangle XYZ is XY, which has a length of 12, and the height is YZ, which has a length of 6. Therefore, the area of triangle XYZ is:

Area(XYZ) = (1/2) * 12 * 6 = 36

Now, we need to find the probability that the area of triangle XYD is at most 20. Since point D is chosen randomly within triangle XYZ, any point within the triangle is equally likely to be chosen. So, we can consider the ratio of the areas. **link deleted by Melody**

To find the probability, we compare the area of triangle XYD to the area of triangle XYZ. The probability is given by:

Probability = Area(XYD) / Area(XYZ)

To find the area of triangle XYD, we need to determine the base and height of this triangle. The base is still XY, which is 12. The height can be found by considering that the area of triangle XYD is half the area of rectangle XYZD.

Since XYD is a right triangle, the height is the altitude from the right angle to the hypotenuse YD. We can use the formula for the area of a triangle to find the height:

Area(XYD) = (1/2) * base * height

20 = (1/2) * 12 * height

height = (2 * 20) / 12 = 40 / 12 = 10 / 3 ≈ 3.33

Therefore, the height of triangle XYD is approximately 3.33.

Now we can calculate the area of triangle XYD:

Area(XYD) = (1/2) * 12 * 3.33 ≈ 19.98

Now we can calculate the probability:

Probability = Area(XYD) / Area(XYZ) = 19.98 / 36 ≈ 0.555

So the probability that the area of triangle XYD is at most 20 is approximately 0.555, or 55.5%.

Georgemichels May 12, 2023