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?
We can start by using the Pythagorean theorem to find the length of the hypotenuse XZ of triangle XYZ:
XZ^2 = XY^2 + YZ^2 = 12^2 + 6^2 = 144 + 36 = 180
Taking the square root of both sides, we get XZ = 6√5.
The area of triangle XYZ is (1/2) * XY * YZ = 36, and the area of rectangle XYX'Z, where X' is the midpoint of XZ, is XY * X'Z = 36√5.
Since triangle XYD is a right triangle with legs of length 6 and a and hypotenuse XY, its area is (1/2) * 6 * a = 3a, where a is the length of the altitude from D to XY.
If we choose D randomly within triangle XYZ, the probability that it lies within a certain region of the triangle is proportional to the area of that region. Therefore, we can find the probability that the area of triangle XYD is at most 20 by finding the ratio of the area of the region where this is true to the area of triangle XYZ.
Let H be the foot of the altitude from X to YZ, and let E be the point where the line parallel to YZ through D intersects XY. Then, triangle XYD has area at most 20 if and only if DE is at most 10.
The area of region XYZE is (1/2) * YZ * XE = 3XE, and the area of region XHY is (1/2) * XY * YH = 36/5. Therefore, the area of the region where the area of triangle XYD is at most 20 is:
36 - 3XE - (36/5) = 144/5 - 3XE
We want this area to be proportional to the area of triangle XYZ, which is 36, so we want:
(144/5 - 3XE)/36 = P
where P is the probability we are looking for.
Solving for XE, we get:
XE = (144/5 - 36P)/3
We need XE to be at most 6 in order for DE to be at most 10. Therefore, we have:
(144/5 - 36P)/3 ≤ 6
Multiplying both sides by 3, we get:
144/5 - 36P ≤ 18
Subtracting 144/5 from both sides, we get:
-36P ≤ -126/5
Dividing both sides by -36 and reversing the inequality, we get:
P ≥ 7/60
Therefore, the probability that the area of triangle XYD is at most 20 is at least 7/60.