# In the given figure ABCD is a trapezium in which AB‖DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar (ABFE) : ar(EFCD) = ?A. A:bB. (a + 3b):(3a + b)C. (3a + b):(a + 3b)D. (2a + b):(3a + b)

Given: ABCD is a trapezium, ABDC, AB = a cm and DC = b cm, E and F are the midpoints of AD and BC.

Since E and F are midpoints of AD and BC, EF will be parallel to both AB and CD.

EF =

Height between EF and DC and height between EF and AB are equal, because E and F are midpoints OF AD and BC and EF||AB||DC.

Let height between EF and DC and height between EF and AB be h cm.

Area of trapezium = 1/2 × (sum of parallel lines) × height

Now,

Area (Trap.ABFE) = 1/2 × (a + ) × h.

and

Area (Trap.ABFE) = 1/2 × (b + ) × h.

Area (Trap.ABFE) : Area (Trap.ABFE) = 1/2 × (a + ) × h : 1/2 × (b + ) × h

Area (Trap.ABFE) : Area (Trap.ABFE) = : = 3a + b : a + 3b

Area (Trap.ABFE) : Area (Trap.ABFE) = 3a + b : a + 3b

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