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# Probability

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64
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I need the answer to this

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?

Mar 5, 2023

#1
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Let's start by drawing a diagram of the triangle XYZ:

```
X
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|          \
|__________\
Y    12     Z
6
```

The area of triangle XYZ is (1/2) * 12 * 6 = 36.

Let's choose a coordinate system with the origin at point X, and the x-axis along the line XY. Then the coordinates of points Y and Z are (12, 0) and (0, 6), respectively.

Since point D is chosen at random within the triangle, any point in the triangle is equally likely to be chosen. So we can assume that the coordinates of point D are (x, y), where 0 ≤ x ≤ 12 and 0 ≤ y ≤ 6.

The area of triangle XYD is given by the formula:

(1/2) * base * height

where the base is the length of XY, and the height is the perpendicular distance from point D to line XY. The height can be expressed as:

h = (6 - y) * (x/12)

Therefore, the area of triangle XYD is:

A = (1/2) * 12 * (6 - y) * (x/12) = x * (6 - y)

We want the probability that the area of triangle XYD is at most 20. So we need to find the region in the xy-plane where x * (6 - y) ≤ 20.

If we graph this region, we get a trapezoid with vertices at (0, 0), (0, 6), (20/3, 2/3), and (12, 0):

```
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|__________\
```

The area of this trapezoid is:

(1/2) * (6) * (20/3) + (1/2) * (12 - 20/3) * (6 - 2/3) = 44/3

Therefore, the probability that the area of triangle XYD is at most 20 is:

P(A ≤ 20) = (area of trapezoid) / (area of triangle XYZ) = (44/3) / 36 = 11/27

So the probability that the area of triangle XYD is at most 20 is 11/27.

Mar 5, 2023