Question 25382: This problem is under Chapter 5, pg. 194 #32 in a Houghton Mifflin Math book. It is really hard,and I am desperate for help.
There is a figure drawn with Trapezoid ABCD with Median MN. Then there are diagonals AC and DB with the Intersecting points, E and F, respectively. The problem is: If DC=3x, AB=2xsquared, and EF=7, find the value of X.
Please Help!!!!!!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! There is a figure drawn with Trapezoid ABCD with Median MN. Then there are diagonals AC and DB with the Intersecting points, E and F, respectively. The problem is: If DC=3x, AB=2xsquared, and EF=7, find the value of X.
THOUGH , I DONOT HAVE YOUR BOOK ,YOU SHOULD BE HAVING IT.SO FOLLOW THE DIAGRAM FROM THE BOOK.I SHALL FOLLOW THE SAME NOMENCLATURE GIVEN BY YOU.O.K?
ABCD IS THE TRAPEZIUM.LET AB BE THE BOTTOM SIDE AND DC THE TOP SIDE.M IS THE MID POINT OF AD,AND N IS THE MID POINT OF BC.MN IS THE MEDIAN.JOIN AC AND BD THE DIAGONALS.LET AC CUT MN AT E AND BD CUT MN AT F.WE ARE GIVEN
EF=7
AB=2X^2
DC=3X
NOW MEDIAN IN A TRAPEZIUM IS PARALLEL TO THE 2 PARALLEL SIDES AB AND DC AND IS HALF THE SUM OF THE 2 PARALLEL SIDES .THAT IS
MN = (AB+DC)/2=(2X^2+3X)/2
AGAIN ,WE HAVE 3 PARALLEL LINES AB,DC AND MN.ONE TRANSVERSAL AD CUTS THEM IN EQUAL PARTS.(AM=MD).HENCE ANY OTHER TRANSVERSAL ALSO CUTS THEM IN EQUAL PARTS.
HENCE FOR TRANSVERSAL DFB WE HAVE
DF=FB
AND FOR TRANSVERSAL CEA WE HAVE
CE=EA.
NOW IN TRIANGLE DAB , MF IS THE LINE JOINING MIDPOINTS OF DA AND DB.HENCE
MF=AB/2=2X^2/2=X^2....
SIMILARLY IN TRIANGLE CAB , EN IS THE LINE JOINING MIDPOINTS OF CA AND AB.HENCE
EN=AB/2=2X^2/2=X^2.......NOW
MN=(2X^2+3X)/2=MF+FN=MF+EN-EF= X^2+X^2-7 =2X^2-7....OR
2X^2+3X=2(2X^2-7)=4X^2-14
4X^2-14-2X^2-3X=0
2X^2-3X-14=0
2X^2-7X+4X-14=0
X(2X-7)+2(2X-7)=0
(2X-7)(X+2)=0
2X-7=0
X=7/2=3.5
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