Point X is randomly selected in square ABCD. Find the probability that this point belongs to the triangle ADM

| March 26, 2021 | education

Point X is randomly selected in square ABCD. Find the probability that this point belongs to the triangle ADM, where point M: a) is the middle of the side CD b) divides the segment CD in the ratio 1: 2 counting from point C c) divides the segment CD in the ratio n: m. counting from point C.

The probability that this point X belongs to the ADM triangle will be equal to the ratio of the areas of the ADM triangle and the square ABCD.

1) Let the side of the square be equal to a cm.

Then Savsd = a ^ 2 cm2.

The KM segment is the middle of the square. AM is the diagonal of the quadrangle ADMK, then Sadmk = Savsd / 2 = (a / 2) * a = a ^ 2/2.

Sadm = Sadmk / 2 = a2 / 4

P = (a ^ 2/4) / a ^ 2 = 0.25.

2)

Point M divides the CD side into two segments DM = 2 * CD. CD = DM / 2.

DM + CD = a.

DM + DM / 2 = a.

3 * DM / 2 = a.

DМ = 2 * a / 3.

Then Sadm = (a * 2/3) * a = a ^ 2 * 2/3.

P = (a ^ 2 * 2/3) / a ^ 2 = 2/3.

3) Point M divides CD into segments CM = n * a / (n + m), DM = m * a / (m + n).

Then Sadm = m * a ^ 2 / (m + n).

P = m * a ^ 2 / (m + n) / a ^ 2 = m / (m + n).

Answer: 1/4; 2/3; m / (m + n).



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