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ABCD is a trapezium in which AB∥DC and AB = 2CD. The diagonals AC and BD meet at O. The ratio of areas of triangles AOB and COD is _______.

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a

1:4

b

1:2

c

2:1

d

4:1

answer is D.

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Detailed Solution

In trapezium ABCD,
AB is parallel to CD and AB = 2 CD ---------- (1)
Diagonals AC and BD intersect at ‘O’
In ΔAOB and ΔCOD,
∠AOB = ∠COD (vertically opposite angles)
∠ABO = ∠CDO (alternate interior angles)
Or, ΔAOB ~ ΔCOD (AA criterion)
Or,

\frac{Area of ΔAOB}{Area Of ΔCOD}

=

\frac{(AB) 2}{(CD) 2}

Or,

\frac{(2 CD) 2}{(CD) 2}

=

\frac{4 CD 2}{(CD) 2}

=

\frac{4}{1}

[From equation (1)]
Thus, Area of ΔAOB : Area of ΔCOD = 4:1
Hence option (4) is correct.

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