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?

Guest Feb 17, 2023

#1**0 **

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.

Guest Feb 17, 2023