Question

ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. A point is selected at random in the square. Calculate the probability that it will liein the triangle MCN.

Answer 1

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:

$A_t = \frac{0.5x*0.5x}{2}=\frac{x^2}{8}$

The area of the square ABCD is:

$A_s =x*x =x^2$

Thus, the probability that a random point lies in the triangle MCN is:

$P=\frac{A_t}{A_s}=\frac{\frac{x^2}{8}}{x^2}=\frac{1}{8}=0.125$

The probability that the point will lie in the triangle MCN is 1/8 or 0.125.