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Quadrilateral $PQRS$ is a trapezoid with bases $\overline{PQ}$ and $\overline{RS}$.  The median $\overline{MN}$ meets the diagonals $\overline{PR}$ and $\overline{QS}$ at $X$ and $Y$, respectively.  If $PS = QR$, then find $RY$.

 

 Jan 5, 2024
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Let the intersection of the  diagonals  =  T

 

The distance from  the  base to  the midline =  5

 

Triangle   PTQ is similar to triangle RTS  and both are isosceles with PT =QT   and ST = RT

So....the height of each  bisect their bases

 

Since  PQ / SR =  30 :20  = 3 : 2.....there are 5 equal parts to the height  of the trapezoid and the height  of triangle PTQ is 3 of them.....so its height =(3/5)10  =  6    and triangle RTS has a height of 4

 

And triangle XTY  is similar to  triangle PTQ

 

So XTY is also isoceles  and its  height also bisects its  base

 

The height of   triangle XTY = height of triangle PTQ - height of midline = 6 - 5  = 1

 

So height of triangle   XTY to height of triangle PTQ =  1 : 6

 

So the base of triangle XTY to the base of triangle PTQ = 1 : 6

 

So the base of triangle XTY = 30 /6 =  5

 

 

Let the midpoint of the base of triangle XTY =  W

 

And   WY = 1/2 the base of triangle XTY  = 5/2 =   2.5

 

Let  W  = (0,0)

Let Y =(2.5,0)

Let  R  = (10,5)

 

So  RY  = sqrt [ (10 - 2.5)^2 + 5^2 ] = sqrt[ 7.5^2 + 25]  = sqrt [ 56.25 + 25 ] =  sqrt [ 81.25]  ≈  9.01

 

 

cool cool cool

 Jan 5, 2024

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