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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

Best Answer 

 #1
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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
avatar
+1
Best Answer

\(\triangle XYD\) 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 :)

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

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