In trapezium ABCD, AB is parallel to CD and the diagonals AC and BD intersect at point O. The longer base AB is twice the shorter base CD (AB = 2 CD). What is the ratio of the area of triangle AOB to the area of triangle COD?

Introduction / Context:
In this geometry question we work with a trapezium, a quadrilateral that has exactly one pair of parallel sides. The problem focuses on the relationship between the parallel sides of the trapezium and the areas of two triangles formed by its diagonals. Understanding how diagonals intersect inside a trapezium and how areas depend on base lengths and heights is essential for solving many coordinate and Euclidean geometry questions in competitive exams.

Given Data / Assumptions:

• ABCD is a trapezium with AB parallel to CD.
• The diagonals AC and BD intersect at point O inside the trapezium.
• The longer base AB is twice the length of the shorter base CD, so AB = 2 * CD.
• We are asked to find the ratio of areas: area(ΔAOB) : area(ΔCOD).
• All points lie in a plane and the figure is a standard Euclidean trapezium.

Concept / Approach:
The key concept is that when diagonals of a trapezium intersect, triangles formed between one base and the intersection point share the same height but have different bases along the parallel sides. For triangles with the same altitude, the ratio of their areas equals the ratio of their corresponding bases. In this problem, the area of triangle AOB is related to base AB, and the area of triangle COD is related to base CD, after we analyze how the diagonals intersect and form similar or proportionate triangles inside the trapezium.

Step-by-Step Solution:
Step 1: Consider trapezium ABCD with AB parallel to CD and AB = 2 * CD. Step 2: Draw the diagonals AC and BD so that they intersect at point O inside the trapezium. Step 3: Use the known property of a trapezium: the diagonals intersect in such a way that the areas of triangles formed near the parallel sides are proportional to the squares of the corresponding bases, or equivalently, that the segments of diagonals relate to the bases. Step 4: A convenient method is to place the trapezium on a coordinate plane. Let AB be the longer base on the x-axis and CD be the shorter base above it. A coordinate model shows that area(ΔAOB) becomes proportional to AB * some height, while area(ΔCOD) becomes proportional to CD * a related height. Step 5: More directly, using the known result derived from coordinate geometry or vector methods, when AB is twice CD, the intersection of diagonals divides the figure so that area(ΔAOB) : area(ΔCOD) = (AB / CD)^2 = (2 / 1)^2 = 4 : 1. Step 6: Therefore the required ratio of areas of triangles AOB and COD is 4 : 1.

Verification / Alternative check:
An easy verification is to assume specific coordinates. For example, take AB from (0,0) to (2,0) and CD from (0,1) to (1,1), creating a simple trapezium with AB = 2 and CD = 1. Compute the intersection point of diagonals AC and BD, then use the coordinate formula for the area of a triangle. When you do this, you will find that area(ΔAOB) is four times area(ΔCOD), which confirms the theoretical ratio 4 : 1 found by the algebraic argument.

Why Other Options Are Wrong:
Option 1: 1:4 is the inverse of the correct ratio. It would suggest that the triangle near the shorter base has a larger area, which contradicts the effect of AB being longer than CD. Option 3: 2:1 ignores the fact that the relationship involves the square of the base ratio, not just the simple linear ratio of the bases. Option 4: 3:1 has no geometric justification from the properties of intersecting diagonals in a trapezium where AB = 2 * CD. Option 5: None of these is incorrect because one of the provided ratios, 4:1, matches the correct area ratio.

Common Pitfalls:
Students often confuse the ratio of areas with the ratio of sides directly, forgetting that in many configurations the area ratio involves squares or products of sides. Another common mistake is to assume that diagonals of a trapezium behave the same way as diagonals in a parallelogram, which is not true. Failing to distinguish between which triangles share a common height and which triangles share a common base can easily lead to an incorrect ratio. It is also easy to accidentally invert the ratio if you do not keep consistent order, such as writing area(ΔCOD) : area(ΔAOB) instead of the ratio requested.

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