How do I deal with this probability
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?
+163 This problem has already been answered :)
https://web2.0calc.com/questions/cphill-help-pls
if that is incorrect I can come back and attempt to recomplete a solution
+2666 Note that that \(\triangle XYD\) has a base of 12. This means that the height of the triangle can be at most 10/3.
We want to find the shaded region's area over the triangle's area, which is 36.
The area of the shaded region is the same as the area of the triangle minus the area of the non-shaded region.
The nonshaded region is a right triangle similar to \(\triangle XYZ\) with a scale factor of 4/9, so its area is \(36 \times ({4 \over 9})^2 = {64 \over 9}\).
This means that the shaded region has an area of \(36 - {64 \over 9} = {260 \over 9}\), so the probability is just \({{260 \over 9} \over {36}} = \color{brown}\boxed{65 \over 81}\)
