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 $12$?

Answer 1

See attached sketch. Suppose the triangle is placed in the first quadrant of the coordinate plane, so the vertex y is at the origin, x is at the point (12, 0), and z is at the point (0, 6). Then the hypotenuse xz is contained in the line (12, 0) and (0, 6), with slope

`(0-6)/(12-0)=-\frac12`

and hence equation

`y - 6 = -\frac12 x \implies y = 6 - \frac x2`

Since ∆xyd and ∆xyz have the leg xy in common, we'll let that be the base. Then the area of the ∆xyd is entirely determined by the vertical distance between the point d and the leg xy. ∆xyz has area 1/2 × 6 × 12 = 36, and so if we let d = (x', y'), we observe x' and y' are jointly distributed with density

`f_(X',Y')(x',y') = \begin{cases}\frac1{36} & \text{if }0 < x' < 12 \text{ and } 0 < y' < 6-\frac{x'}2 \\ 0 & \text{otherwise}\end{cases}`

The area of ∆xyd is

`\frac12 * 12 * y' = 6y'`

and so we want to find

`\mathrm{Pr}\left\{6y' \le 12\right\} \iff \mathrm{Pr}\left\{y' \le 2\right\}`

Now, the event that y' ≤ 2 can be split up into cases of

• 0 < x' ≤ 8 and 0 < y' ≤ 2

• 8 ≤ x' < 12 and 0 < y' < 6 - x'/2

So we have

`\displaystyle \mathrm{Pr}\left\{y'\le2\right\} = \int_0^8 \int_0^2 f_(X',Y')(x',y') \, dy' \, dx' + \int_8^(12) \int_0^{6-\frac{x'}2} f_(X',Y')(x',y') \, dy' \, dx'`

`\displaystyle \mathrm{Pr}\left\{y'\le2\right\} = \frac1{36} \int_0^8 \int_0^2 dy' \, dx' + \frac1{36} \int_8^(12) \int_0^{6-\frac{x'}2}dy' \, dx'`

`\displaystyle \mathrm{Pr}\left\{y'\le2\right\} = \frac49 + \frac19 = \boxed{\frac59}`

(i.e. the area of a trapezoid with base lengths 8 and 12 and height 2, divided by the total area of ∆xyz)