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# Probability

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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 10?

Feb 15, 2024

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$$\triangle XYD$$ has base $$XY=12$$ and Hmax of:

$$\frac{1}{2}bh=10\\ h=\frac{5}{3}$$
Thus the desired area is $$\frac{5}{3}$$ from base $$XY$$ (A straight line parallel to $$XY$$)
Let the line touch $$YZ$$ at $$A$$ and $$ZX$$ at $$B$$
$$P=\frac{\text{desired area}}{\text{total area}}$$

So: Total area=$$\frac{1}{2}bh=36$$ and desired area is:

$$\frac{1}{2}h(XY+AB)=\frac{155}{9}$$  ($$AB=\frac{26}{3}$$ due to similar triangles $$\triangle ZAB \sim \triangle ZYX$$)

Thus: $$\boxed{P=\frac{155}{324}}$$

ggwp :)

Feb 15, 2024
edited by EnormousBighead  Feb 15, 2024
edited by EnormousBighead  Feb 15, 2024

#1
+1

$$\triangle XYD$$ has base $$XY=12$$ and Hmax of:

$$\frac{1}{2}bh=10\\ h=\frac{5}{3}$$
Thus the desired area is $$\frac{5}{3}$$ from base $$XY$$ (A straight line parallel to $$XY$$)
Let the line touch $$YZ$$ at $$A$$ and $$ZX$$ at $$B$$
$$P=\frac{\text{desired area}}{\text{total area}}$$

So: Total area=$$\frac{1}{2}bh=36$$ and desired area is:

$$\frac{1}{2}h(XY+AB)=\frac{155}{9}$$  ($$AB=\frac{26}{3}$$ due to similar triangles $$\triangle ZAB \sim \triangle ZYX$$)

Thus: $$\boxed{P=\frac{155}{324}}$$

ggwp :)