Medians $\overline{DP}$ and $\overline{EQ}$ of $\triangle DEF$ are perpendicular. If $DP= 18$ and $EQ = 24$, then what is ${DF}$?
+128475 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
