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Medians $\overline{DP}$ and $\overline{EQ}$ of $\triangle DEF$ are perpendicular. If $DP= 18$ and $EQ = 24$, then what is ${DF}$?

Guest Mar 25, 2018
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Let R be the intersection of the medians

DR  =1/3  of DP  = 6

ER  =1/3  of EQ  = 8

So....since the medians are perpendicular....then  DE  is the hypotenuse of right triangle DRE  

So....call S  the midpoint  of DE...so  DS  = 5

So.... SF  is another median

The sine of RDE  = 8/10  = 4/5

cos RDE   = 3/5

 

 

 

So the distance  from  R to S  can be found using the Law of Cosines  as

RS^2  = DS^2 + DR^2 -2(DS * DR)cos(RDE)

RS^2  = 5*2 + 6^2  - 2(30)cosRDE

RS^2  = 61 - 60(3/5)

RS^2  = 61  - 36

RS^2  = 25

RS  = 5

 

But  RS  = 1/3  of SF....so SF  = 15

 

We need to find  sin DSR

 

Using the Law of Sines we have

 

sinDSR / DR  = sinRDE/ RS

sinDSR/6  = sinRDE / 5

sinDSR / 6   = (4/5) / 5

sinDSR / 6  = 4/25

sinDSR  = 24/25

 

And the cos DSR   = √ [1  - (24/25)^2 ]  = √ [625 - 576 ] / 25  = √49 / 25 =  7/25

 

So...we can find DF with the Law of Cosines as

 

DF^2  = DS^2  + SF^2  - 2(DS*SF)cos(DSR)

DF^2  = 5^2  + 15^2  - 2(5*15)(7/25)

DF^2  = 250  - 2*3*7

DF  = 250 - 42

DF^2  = 208

 

DF = √208  = 4√13

 

 

cool cool cool

CPhill  Mar 25, 2018

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