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 $10$?
The area of triangle $XYZ$ is $\frac12(12)(6)=36$, so $XD \cdot YD \le 80$. We can use this to write inequalities showing where point $D$ cannot be: if $XD > 10$, then the area of triangle $XYD$ is at least $\frac12(12)(10)=60$ (and similarly if $YD > 10$), and if $XD < 5$, then $YD > \frac{80}{XD} > 16$. Thus, the probability that $XD \le 10$ and $YD \le 10$ is the same as the probability that $5 \le XD \le 10$ and $5 \le YD \le 10$: this is the desired area divided by 36, or $\frac{(10-5)(10-5)}{36} = \boxed{\frac{25}{36}}$.
To find the probability that the area of triangle $XYD$ is at most $10$, we need to compare the area of triangle $XYD$ with the total area of triangle $XYZ$.
First, let's find the area of triangle $XYZ$. The area of a triangle can be calculated using the formula:
Area = $\frac{1}{2} \times$ base $\times$ height
In this case, the base $XY$ is given as 12, and the height can be found by dropping a perpendicular from vertex $Z$ to side $XY$. Since triangle $XYZ$ is a right triangle, this perpendicular will bisect side $XY$. Therefore, the height is half the length of $YZ$, which is 3.
So the area of triangle $XYZ = \frac{1}{2} \times 12 \times 3 = 18$.
Now, let's find the area of triangle $XYD$. We know that the two legs of this triangle are $XY = 12$ and $YD = $ (some length between 0 and 6). To find the area, we can use the same formula:
Area of triangle $XYD = \frac{1}{2} \times$ base $\times$ height
In this case, the base $XY$ is still 12, and the height is $YD$. To get a bound on $YD$ that satisfies the condition 'at most 10', we need to find the largest possible height when the area is exactly 10.
Let's assume the height is $h$. The area of triangle $XYD$ is then $\frac{1}{2} \times 12 \times h$. We want this to be 10:
$\frac{1}{2} \times 12 \times h = 10$
Simplifying, we find $h = \frac{10}{6} = \frac{5}{3}$.
So, the height $YD$ should be less than or equal to $\frac{5}{3}$.
Now, we have all the information needed to find the probability:
Probability = $\frac{\text{Area of triangle }XYD \text{ is at most 10}}{\text{Area of triangle }XYZ}$
Substituting the values we found:
Probability = $\frac{\frac{1}{2} \times 12 \times \frac{5}{3}}{18}$
Simplifying further, we get:
Probability = $\frac{20}{18} = \frac{10}{9} \approx 1.11$
Therefore, the probability that the area of triangle $XYD$ is at most 10 is approximately $1.11$, or $\frac{10}{9}$.
Given that triangle $XYZ$ is a right triangle, we can use the formula for the area of a triangle:
$$\text{Area} = \frac12 \times \text{base} \times \text{height}$$
Let's call the height of triangle $XYD$ as $h$. We know that the base $XY$ has a length of $12$. Since $D$ is chosen at random within triangle $XYZ$, the position of $D$ can be anywhere within the triangle. Therefore, the height $h$ can vary from $0$ (when $D$ is on the line segment $XY$) to $6$ (when $D$ is on the line segment $YZ$).
To find the probability that the area of triangle $XYD$ is at most $10$, we need to find the portion of the possible values for $h$ that satisfies this condition.
Notice that the area of triangle $XYD$ is $\frac12 \times 12 \times h = 6h$. So we are looking for $h$ values that make $6h \leq 10$.
Solving this inequality, we have:
$$h \leq \frac{10}{6} = \frac{5}{3}$$
Since $h$ can vary from $0$ to $6$, the portion of possible $h$ values that satisfy the condition is $\frac{5}{3}$ out of $6$. Therefore, the probability is:
$$\frac{\text{portion that satisfies condition}}{\text{total portion}} = \frac{\frac{5}{3}}{6} = \frac{5}{18}$$
So the probability that the area of triangle $XYD$ is at most $10$ is $\frac{5}{18}$.