Question

ABCD is a trapezium in which AB||DC,AB=16 cm and DC=24 cm.If E and Fa re respectively the midpoints of Ad andf BC,prove that $ar\left(ABFE\right)=\frac{9}{11}ar\left(EFCD\right)$.

Answer 1

Let E be the mid point of AD and F be the mid point of CB.
Const : Join EF, AC and construct a perpendicular from A on CD and mark the point on CD as G and the point on AG intersecting EF as H.

Now, In triangle ADC, let O, the intersection point of AC and EF, be the mid point of AC
Thus, EO // CD and EO = 1/2 CD = 12 cm
Similarly, OF = 1/2 AB = 8 cm
So, EF = OE + OF = 12+8 = 20 cm

In triangle ADG,
EH // CD and E is the mid point.
Therefore, H is also the mid point of AG (converse MPT)
AH = GH----------(i)

Now we compare the area of the trapeziums
ar(DCEF) = (\frac{1}{2}(20+24) x AH : ar(EFBA) = 1/2(20+16) x GH
as AH = GH,
= \frac{1}{2} x 44 = 1/2 x 36 9/11
= 22 : 18
= 11 : 9
Hence Prove ar(ABFE)=9/11 ar(EFCD).