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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?
Guest Mar 16, 2023
+4 Let $A$ denote the area of triangle $XYZ$. We find that $A = (1/2)(12)(6) = 36$. Since the area of triangle $XYD$ is half the area of triangle $XYZ$, triangle $XYD$ has area at most 20 if and only if $XYD$ has area at most 10.
Let $E$ denote the foot of the altitude from $X$ to line $DZ$. Then triangle $XYE$ is similar to triangle $XYZ$, so the altitude from $X$ to $YZ$ has length $3$. Since $XY=12$, the length of the altitude from $X$ to line $DY$ is $2A/XY=3/2$.
Hence the length of $DE$ is $6-(3/2)=9/2$, so the area of triangle $XYD$ is $(1/2)(12)(9/2)=27$, which is greater than $20$. Therefore, the probability that the area of triangle $XYD$ is at most 20 is $\boxed{0}$.
+4 We can begin by drawing a diagram of triangle XYZ with legs XY = 12 and YZ = 6. Let's label the right angle as angle Z, and the hypotenuse as XZ.
Next, let's draw a line segment from point D perpendicular to leg XY, and call the point of intersection E. We can see that triangle XYD is now divided into two smaller triangles, XYE and EYD.MyBalanceNow
Since we know the lengths of legs XY and YZ, we can use the Pythagorean theorem to find the length of hypotenuse XZ. XZ^2 = XY^2 + YZ^2, so XZ^2 = 12^2 + 6^2 = 144 + 36 = 180.
