SOLUTION: The parallel sides of trapezoid ABCD are 3 cm and 9 cm(AB and DC).The non parallel sides are 4 cm and 6 cm(AD and BC).A line(PQ) parallel to base(DC) divides the trapezoid into tw

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Question 363490: The parallel sides of trapezoid ABCD are 3 cm and 9 cm(AB and DC).The non parallel sides are 4 cm and 6 cm(AD and BC).A line(PQ) parallel to base(DC) divides the trapezoid into two trapezoids(ABQP and PQCD ) of equal perimeters. Find the ratio in which each of the non parallel sides is divided,if AP:BQ=PD:QC
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Note: the word is "perimeter", not "parameteras", as you had it.
"Perimeter" means the distance all the way around a figure.

I corrected that :-)

The parallel sides of trapezoid ABCD are 3 cm and 9 cm(AB and DC).The non parallel sides are 4 cm and 6 cm(AD and BC).A line(PQ) parallel to base(DC) divides the trapezoid into two trapezoids(ABQP and PQCD ) of equal perimeters. Find the ratio in which each of the non parallel sides is divided,if AP:BQ=PD:QC
Here's the trapezoid:



Now we'll draw in PQ parallel to the top and bottom of the trapezoid,
cutting it into two trapezoids with equal perimeters. 


 
Now I'll explain my labeling of those parts:

AP%2FBQ+=+PD%2FQC

AP%2AQC+=+PD%2ABQ

Divide both sides by PD%2AQC

%28AP%2AQC%29%2F%28PD%2AQC%29+=+%28PD%2AQC%29%2F%28PD%2AQC%29



AP%2FPD+=+BQ%2FQC

Take reciprocals of both sides

PD%2FAP+=+QC%2FBQ

Add 1 to both sides

PD%2FAP+%2B+1=+QC%2FBQ%2B1

Replace 1 by AP%2FAP on the left and replace 1 on the right by BQ%2FBQ

PD%2FAP+%2B+AP%2FAP=+QC%2FBQ+%2B+BQ%2FBQ

Combine numerators over denominators:

%28PD%2BAP%29%2FAP+=+%28BQ%2BQC%29%2FBQ

Since AP+PD = AD = 6 and BQ+QC = BC = 4

6%2FAP+=+4%2FBQ

Take reciprocals of both sides

AP%2F6+=+BQ%2F4

Let that ratio be = k

AP%2F6+=+BQ%2F4+=+k

AP+=+6k and BQ=4k

BQ = 4k,
QC = 4 - 4k
AP = 6k
PD = 6 - 6k 
 
all in centimeters. 
 
Perimeter of the upper trapezoid ABQP:

BQ + AB + AP + PQ

Perimeter of the lower trapezoid PQCD:

QC + DC + PD + PQ

Equating the two perimeters:

BQ + QB + AP + PQ = QC + DC + PD + PQ

Subtract PQ (their common side) from both sides:

     BQ + QB + AP = QC + DC + PD

Substitute their lengths in terms of the fraction k:
 
      4k + 3 + 6k = (4-4k) + 9 + (6-6k)

          10k + 3 = 4 - 4k + 9 + 6 - 6k

          10k + 3 = 19 - 10k
      
              20k = 16

                k = 16%2F20 = 4%2F5 = 0.8

Therefore

AP = 6k = 6(0.8) = 4.8 cm.  

BQ = 4k = 4(0.8) = 3.2 cm. 

PD = 6-6k = 6-6(0.8) = 1.2 cm.

QC = 4-4k = 4-4(0.8) = 0.8 cm

We want to find

AP:PD = BQ:QC

which is 4.8:1.2 = 48:12 = 4:1

and checking:

BQ:QC = 3.2:0.8 = 32:8 = 4:1

That's the answer 4:1

Incidentally, the figures above are drawn approximately to scale.
Measure them with a ruler and you will see they are fairly close to
3cm, 4cm, 9cm and 6cm.
  
The perimeters are equal even though the upper trapezoid is obviously 
lots bigger in area.  This demonstrates the important fact that two
figures can have the same perimeter and quite different areas.  

Edwin