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 $12$?

Guest Jul 29, 2022

#4**+1 **

Wouldn't the probability be the \({\text{successful region} \over \text{total region}} = {[XYAB] \over \triangle XYZ}\)

The area of the successful region is \((2 \times 8) + (4 \times 2 \div 2) = 20\), and the area of the total region is \(12 \times 6 \div 2 = 36\)

So the probability is \({20 \over 36} = \color{brown}\boxed{5 \over 9}\)

BuilderBoi Jul 29, 2022

#1**-1 **

In order for the area of triangle XYD to be at most 12, the height of triangle XYD is at most 2. The height of triangle XYD can be 6, so the probability is 2/6 = 1/3.

Guest Jul 29, 2022

#3**+3 **

XY = the base of the triangle = 12

Then the altitude cannot be more than 2

After some reflection, I believe that BuilderBoi's answer is correct !!!

D can fall ANYWHERE in the area of the trapezoidal area given by

(1/2) (2) (8 + 12) = 20

So.....the probability is 20 / 36 = 5/9

THX BuilderBoi for helping me to see this !!!!

CPhill Jul 29, 2022

#4**+1 **

Best Answer

Wouldn't the probability be the \({\text{successful region} \over \text{total region}} = {[XYAB] \over \triangle XYZ}\)

The area of the successful region is \((2 \times 8) + (4 \times 2 \div 2) = 20\), and the area of the total region is \(12 \times 6 \div 2 = 36\)

So the probability is \({20 \over 36} = \color{brown}\boxed{5 \over 9}\)

BuilderBoi
Jul 29, 2022

#5**+3 **

a wise owl told me once

In order for the area of triangle XYD to be at most 12, the height of triangle XYD is at most 2. The height of triangle XYD can be 6, so the probability is 2/6 = 1/3.

WISEOWL's answer, which was voted as a 'best answer'

nerdiest
Jul 29, 2022

#7**+1 **

Right, but the area of the bottom 2 units is NOT 1/3 of the total area of the triangle.

BuilderBoi
Jul 29, 2022