Answer:
1/8 or 0.125
Step-by-step explanation:
Let x be the length of the sides of the square ABCD.
Therefore, the length of CN and CM is x/2.
The area of the triangle MCN is:
[tex]A_t = \frac{0.5x*0.5x}{2}=\frac{x^2}{8}[/tex]
The area of the square ABCD is:
[tex]A_s =x*x =x^2[/tex]
Thus, the probability that a random point lies in the triangle MCN is:
[tex]P=\frac{A_t}{A_s}=\frac{\frac{x^2}{8}}{x^2}=\frac{1}{8}=0.125[/tex]
The probability that the point will lie in the triangle MCN is 1/8 or 0.125.
