- Some BodyLv 72 years agoFavorite Answer
Not sure if there's an easier way, but here's how I did it.
ABCD is a trapezoid, so side CD is parallel to side AB.
Therefore, angle CAB and angle ACD are alternate angles and therefore congruent.
Also, angle APB and angle CPD are vertical angles and therefore congruent.
Since angles of a triangle add up to 180, angle ABP and angle CDP are congruent.
Therefore, triangles ABP and CDP are similar.
That means the height of triangle ABP (I'll call it x) and the height of triangle CDP (I'll call it y) can be related through proportions:
x/y = 30/20
x/y = 3/2
The overall height of the trapezoid is x+y:
h = x + y
h = 3/2 y + y
h = 5/2 y
y = 2/5 h
So line XY is located 2/5 of the way from CD to AB.
Next, triangle ABD and triangle XPD are similar. Also, triangle ABC and triangle PYC are similar.
XP / 30 = 2/5
XP = 12
PY/30 = 2/5
PY = 12
XY = XP + PY
XY = 24
- ted sLv 72 years ago
[ 24 ] units.............let A = (0,0) and D = ( 0 , W ) ; C =(20 , W) ; B= ( 30 , 0) and line BC is y1 = - W / 10 ( x - 30) ;
line BD is y2 = - W / 30 ( x - 30) , line AC is y3 = W / 20 x...find intersection of y2 & y3...use that information in y1;
interesting problem for 7th graders
- PuzzlingLv 72 years ago
The length of the line XY is the harmonic mean of the two bases.
a = 20
b = 30
c = H(a,b)
c = 2ab / (a+b)
c = 2(20)(30) / (20+30)
c = 1200/50
c = 24
XY is 24Source(s): http://donsteward.blogspot.com/2010/10/trapezium-p... https://www.cut-the-knot.org/triangle/HarmonicMean... http://jwilson.coe.uga.edu/EMT725/HM/HM.html