Right triangle XYZ has legs of length XY = 12 and YZ = 10. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 24?

MEMEGOD Jul 14, 2024

#1**0 **

Finding the Area of Triangle XYZ

The area of a right triangle is (base * height)/2.

In this case, base = XY = 12 and height = YZ = 10.

So, area of triangle XYZ = (12 * 10)/2 = 60.

Relating the Area of XYD to the Height from D to XY

Let h be the height of triangle XYD from point D to side XY.

The area of triangle XYD = (base * height)/2 = (12 * h)/2 = 6h.

Finding the Maximum Height for an Area of 24

We want the area of XYD to be at most 24.

So, 6h <= 24.

Solving for h, we get h <= 4.

Calculating the Probability

The probability of choosing a point D with a height less than or equal to 4 is equal to the ratio of the area where the height is less than or equal to 4 to the total area of triangle XYZ.

If we draw a line parallel to XY at a distance of 4 units from XY, it will divide the triangle into two similar triangles. The smaller triangle will have a height of 4 and its area will be (1/4) * (area of triangle XYZ) = (1/4) * 60 = 15.

So, the probability of choosing a point D with a height less than or equal to 4 (and thus an area of XYD at most 24) is 15/60 = 1/4.

Therefore, the probability that the area of triangle XYD is at most 24 is 1/4.

nathanl6656 Jul 14, 2024