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?
First, we can find the hypotenuse of the right triangle using the Pythagorean theorem:
XZ^2 = XY^2 + YZ^2 XZ^2 = 12^2 + 6^2 XZ^2 = 144 + 36 XZ^2 = 180 XZ = sqrt(180) = 6*sqrt(5)
So the area of the triangle XYZ is:
A(XYZ) = (1/2)XYYZ = (1/2)126 = 36
Let the length of the altitude from vertex X to side YZ be h. Then the area of triangle XYD is:
A(XYD) = (1/2)XYh
We can use the fact that the area of a triangle is equal to (1/2) times the base times the height to find the length of the altitude from X to YZ. We have:
A(XYD) <= 20
(1/2)XYh <= 20
h <= 40/XY
h <= 40/12
h <= 10/3
So the length of the altitude from X to YZ is at most 10/3.
The area of triangle XYZ is 36, so the ratio of the area of triangle XYD to the area of triangle XYZ is:
A(XYD)/A(XYZ) = (1/2)XYh/((1/2)XYYZ) = h/YZ
For the area of triangle XYD to be at most 20, we need:
(1/2)XYh <= 20
h <= 40/XY
So we have:
A(XYD)/A(XYZ) = h/YZ <= (10/3)/6 = 5/9
Therefore, the probability that the area of triangle XYD is at most 20 is 5/9.