Question

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?

Answer 1

Explanation:

a probability is always

desired cases / totally possible cases.

now in our case the totally possible area is the area of the triangle XYZ (when D = Z).

the desired case is the area of 20 (which contains also the possibilities of smaller areas).

so, the probability is 20/area.

now, let's find the area of XYZ :

Pythagoras tells us the length of the Hypotenuse :

XZ² = XY² + YZ² = 12² + 6² = 144 + 36 = 180

XZ = sqrt(180)

the area based on all 3 sides we get via Heron's formula

s = (a+b+c)/2

A = sqrt(s(s-a)(s-b)(s-c))

s = (12+6+sqrt(180))/2 = 9 + sqrt(180/4) = 9 + sqrt(45)

sqrt(180) = 2×sqrt(45)

A = sqrt((9+sqrt(45))(-3+sqrt(45))(3+sqrt(45))(9-sqrt(45)))

(a+b)(a-b) = a² - b²

so,

(9+sqrt(45))(9-sqrt(45)) = 81 - 45 = 36

(-3+sqrt(45))(3+sqrt(45)) = 45 - 9 = 36

A = sqrt(36×36) = 6×6 = 36

the probability that the area of XYD is max. 20 is

20/36 = 5/9