Join the diagonal AC such that it cuts the line EF at the point Y
From the figure we know that E and F are the midpoints of AD and BC
So we get
EF || AB || CD
Consider △ ACF
We know that E is the midpoint of AD and EY || CD
So we get Y as the midpoint of AC
It can be written as
EY = ½ CD
Consider △ ABC
We get
FY = ½ AB
We know that
EF = EY + YF
By substituting the values
EF = ½ CD + ½ AB
By taking ½ as common
EF = ½ (CD + AB)
By substituting the values
EF = (24 + 16)/2
So we get
EF = 20cm
Construct AQ ⊥ DC such that AQ cuts EF at P
We know that P is the midpoint of AQ
So we get
AP = PQ = h
We get
Area of trapezium ABFE = 1/2 × (EF + AB) × AP
By substituting the values
Area of trapezium ABFE = ½ × (20 + 16) × h
On further calculation
Area of trapezium ABFE = 18h cm2
Area of trapezium EFCD = ½ × (EF + CD) × PQ
By substituting the values
Area of trapezium EFCD = ½ × (20 + 24) × h
On further calculation
Area of trapezium EFCD = 22h cm2
By division
Area of trapezium ABFE/ Area of trapezium EFCD = 18h/ 22h
So we get
Area of trapezium ABFE/ Area of trapezium EFCD = 9/11
It can be written as
Area of trapezium ABFE = 9/11 × (Area of trapezium EFCD)
Therefore, it is proved that ar (ABFE) = 9/11 ar (EFCD).