Probability

We can solve this problem using geometric probability. Let A be the area of triangle XYZ, which is (1/2)(12)(6) = 36. We can think of the point D as being chosen uniformly at random in the triangle XYZ. This means that any point in the triangle is equally likely to be chosen.

Let's consider the set of all possible points D that could be chosen. Since D can be any point in the triangle, this set is just the triangle XYZ itself. Let's call the set of all points within the triangle that make triangle XYD have area at most 20 as set S.

We want to find the probability that a randomly chosen point D is in set S. This probability is given by:

P(D in S) = (Area of S) / (Area of XYZ)

Let's find the area of S. We can think of S as being the region above line XY and below line DZ, where D is any point on line XY such that triangle XYD has area 20. Let's call the intersection of line XY and line DZ point E. Then triangle XYD has base 12 and height 20/12 = 5/3. This means that point D lies on a line parallel to YZ and 5/3 units above line XY. Let's call this line l.

The point E divides line YZ into two segments of length 2 and 4, since XY is perpendicular to YZ and XY = 12. Therefore, the distance from point E to line YZ is 2/3 of the length of YZ, or 4. We can use this information to find the equation of line l in slope-intercept form:

The slope of line l is -3/5 (since it is parallel to YZ and perpendicular to XY), and it passes through point (0, 5/3) (since it is 5/3 units above XY). Therefore, the equation of line l is:

y = (-3/5)x + 5/3

We want to find the x-coordinates of the points where line l intersects lines XY and DZ. Let's call these points F and G, respectively. To find the x-coordinate of point F, we plug in y = 0 into the equation of line l:

0 = (-3/5)x + 5/3

Solving for x, we get:

x = 5

Therefore, point F has coordinates (5, 0). To find the x-coordinate of point G, we plug in y = 6 into the equation of line l:

6 = (-3/5)x + 5/3

Solving for x, we get:

x = 19/3

Therefore, point G has coordinates (19/3, 6). The area of triangle XYD is (1/2)(12)(DF), where DF is the distance from point D to line XY. We want this area to be at most 20, so we have:

(1/2)(12)(DF) ≤ 20

Solving for DF, we get:

DF ≤ 10/3

This means that any point D on line l such that DF ≤ 10/3 will be in set S. We can find the x-coordinates of these points by solving the inequality:

(-3/5)x + 5/3 ≤ 10/3

Solving for x, we get:

x ≤ 25/3

Therefore, the set S is the region bounded by lines XY, DZ, and the line:

y = (-3/5)x + 5/3

in the triangle XYZ, where the x-coordinate of any point in this region is at most 25/3.

We can find the area of S by finding the area of the region bounded by lines XY, DZ, and the line y = (-3/5)x + 5/3 whose x-coordinate is at most 25/3. We can break this region into two triangles and a trapezoid. The area of the two triangles is:

(1/2)(12)(5/3) + (1/2)(25/3)(6-5/3) = 25/3

The area of the trapezoid is:

(1/2)(25/3 + 5)(5/3) = 100/9

Therefore, the area of S is:

25/3 + 100/9 = 275/27

Finally, we can find the probability that a randomly chosen point D is in set S by dividing the area of S by the area of XYZ:

P(D in S) = (275/27) / 36 = 275/972

Therefore, the probability that the area of triangle XYD is at most 20 is 275/972.