Question

ABCD is a parallelogram such that AB is parallel to
DC and DA parallel to CB. The length of side AB is
20 cm. E is a point between A and B such that the
length of AE is 3 cm. F is a point between points D
and C. Find the length of DF such that the segment
EF divides the parallelogram in two regions with
equal areas.

Answer 1

Solution:-

Let A1 be the AEFD trapeze area. Thus,

》A1 = (1/2) h (AE + DF) = (1/2) h (3 + DF), h is the height of the parallelogram.

Now, let A2 be the area of ​​the EBCF trapeze. Thus,

》A2 = (1/2) h (EB + FC)

We also have

》EB = 20 - AE = 17, FC = 20 - DF

Now we substitute EB and FC in A2 = (1/2) h (EB + FC)

》A2 = (1/2) h (17 + 20 - DF) = (1/2) h (37 - DF)

For EF to divide the parallelogram into two regions of equal areas, we need the A1 area and the A2 area to be equal

》(1/2) h (3 + DF) = (1/2) h (37 - DF)

Multiply both sides by 2 and divide thm by h to simplify

》3 + DF = 37 - DF

Solve for DF

》2DF = 37 - 3

》2DF = 34

》DF = 17 cm

☆i hope its help☆