Let def be an equilateral triangle with side length 3 At random, a point g is chosen inside the triangle. Compute the probability that the length dg is less than or equal to 1
also
A stick has a length of 5 units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are shorter than 3 units?
and
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?
+580 I double-checked 1 myself:
The area of a sector is x/360*pi*r^2
60/360*pi*1
= pi/6
The area of the triangle is 1/2*s1*s2 *sin(60deg)
1/2*3*3 *sin(60deg)
= (9sqrt(3)/4)
The prob is the area of the sector / area of the triangle.
giving us:
\(\frac{2\sqrt{3}\pi }{81}\)
Try the answer above^^^^^
I will probably answer 2 and 3 shortly
+580 Number 3:
https://web2.0calc.com/questions/help-last-answer-was-wrong-please-ignore-dollar-signs
Tell me if that is correct or wrong....
+580 Question 2 is 4/25
As stated in:
https://web2.0calc.com/questions/attempted-twice-i-can-t-find-the-answer
+2668 Problem 3: The height of \(\triangle {XYD} \) must be at most 4.
Shading all the points that work, we can find the area of the shaded region over the area of the entire triangle.
The area of the entire triangle is 36, and the area of the shaded region is 20.
Thus the probability is \({20 \over 36 } = \color{brown}\boxed{5 \over 9}\)
Here is the image:
Note: You will get the same answer if you swap the height and width of the triangle.
