Question

ABCD is a trapezium in which AB || DC, AB = 16cm and DC = 24cm. If E and F are respectively the midpoints of AD and BC, prove that ar (ABFE) = 9/11 ar (EFCD).

Answer 1

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).