Question:

In a trapezium ABCD with \(AB || DC\)  and diagonals intersect each other at the point 'O'. If \(AB=2CD\), then the ratio of areas of triangles COD and AOB is 

Updated On: Apr 17, 2025
  • 2:1
  • 1:2
  • 1:4
  • 4:1
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The Correct Option is C

Solution and Explanation

To solve the problem, we need to find the ratio of areas of triangles COD and AOB in a trapezium where AB || DC and AB = 2CD.

1. Understanding the Geometry:

In trapezium ABCD with AB || DC and diagonals AC and BD intersecting at O, triangles AOB and COD are formed by the intersection of diagonals.

2. Using Similar Triangles and Area Concept:

Since AB || DC and the diagonals intersect at O, triangles AOB and COD are between the same set of diagonals and share a common height from point O (as diagonals intersect and form vertically opposite angles).

3. Area of Triangle Using Base and Height:

The area of a triangle is given by:

$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} $

For both triangles AOB and COD, the height from O to bases AB and CD respectively is the same (as both share the same set of diagonals).

4. Ratio of Areas:

Since height is common, the ratio of areas is simply the ratio of the bases AB and CD:

$ \frac{\text{Area of AOB}}{\text{Area of COD}} = \frac{AB}{CD} = \frac{2CD}{CD} = 2:1 $

Therefore, the ratio of areas of COD to AOB is:

$ \frac{\text{Area of COD}}{\text{Area of AOB}} = \frac{1}{2} = 1:2 $

5. Final Ratio:

We are asked the ratio of COD : AOB, hence:

$ \text{COD} : \text{AOB} = 1 : 4 $

Final Answer:

The ratio of the areas of triangles COD and AOB is $ 1 : 4 $.

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