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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 Apr 7, 2023

#1

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We can start by finding the area of triangle XYZ using the formula:

Area = 0.5 * base * height

We know that the legs of the right triangle are XY = 12 and YZ = 6, so the area of XYZ is:

Area(XYZ) = 0.5 * 12 * 6 = 36

Next, we need to find the range of values for the height of triangle XYD such that the area is at most 20.

Area(XYD) = 0.5 * 12 * h = 6h

We want 6h ≤ 20, so h ≤ 20/6 = 10/3.

Therefore, the height of triangle XYD can range from 0 to 10/3.

The probability that point D is chosen such that the area of triangle XYD is at most 20 is equal to the ratio of the area of the triangle XYD to the area of triangle XYZ.

Area(XYD) = 0.5 * 12 * h = 6h

So, the maximum area of XYD that satisfies the condition is 0.5 * 12 * 10/3 = 20.

Therefore, the probability of the area of triangle XYD being at most 20 is:

P = Area(XYD) / Area(XYZ)

P = 20 / 36

P = 5 / 9

Therefore, the probability that the area of triangle XYD is at most 20 is 5/9.

Guest Apr 8, 2023