In trapezium ABCD, AB is parallel to CD and AB is four times CD. The diagonals AC and BD intersect at point O. What is the ratio of the area of triangle DCO to the area of triangle ABO?

Introduction / Context:
This geometry question tests properties of trapeziums and how diagonals divide areas of triangles inside the figure. The key idea is that when a quadrilateral is a trapezium with parallel sides, its diagonals intersect in such a way that the areas of triangles formed at the ends of the diagonals are related to the lengths of the parallel sides. Understanding these properties helps in solving many competitive exam questions very quickly without doing heavy coordinate geometry for every case.

Given Data / Assumptions:

• ABCD is a trapezium.
• AB is parallel to CD.
• AB = 4 × CD.
• Diagonals AC and BD intersect at point O.
• We are asked to find area(△DCO) : area(△ABO).

Concept / Approach:
In a trapezium with one pair of opposite sides parallel, diagonals divide each other in the same ratio as the lengths of the parallel sides. If AB and CD are parallel and AB > CD, then along each diagonal, the division of segments is in the ratio AB : CD. Further, the area of two triangles that have the same height is proportional to the length of their corresponding bases. When diagonals intersect, triangles at the ends of the trapezium that share the same height but sit on different parallel bases have areas in the ratio of those bases squared, but here we compare triangles placed on different parts of the same diagonals and use the segment ratio result.

Step-by-Step Solution:
Step 1: Let AB = 4k and CD = k, so that AB : CD = 4 : 1.Step 2: In a trapezium with AB ∥ CD, diagonals AC and BD intersect at O such that AO / OC = BO / OD = AB / CD = 4 / 1.Step 3: Therefore AO : OC = 4 : 1, so AO = 4x and OC = x for some x. Similarly BO = 4y and OD = y for some y.Step 4: Consider triangles ABO and DCO. These triangles are similar by the property of vertical angles and parallel sides, but their corresponding linear dimensions along the diagonals are in the ratio AO : OC and BO : OD, both equal to 4 : 1.Step 5: When corresponding linear dimensions of similar triangles are in ratio 4 : 1, the ratio of their areas is 4^2 : 1^2 = 16 : 1.Step 6: Since triangle ABO corresponds to the larger side and triangle DCO to the smaller side, area(△ABO) : area(△DCO) = 16 : 1, so area(△DCO) : area(△ABO) = 1 : 16.

Verification / Alternative check:
An alternative approach is to place the trapezium in a coordinate plane for confirmation. For example, set A = (0, 0), B = (4, 0), D = (0, h) and C = (1, h) so that AB = 4 and CD = 1. Finding the intersection O of lines AC and BD and then computing the areas of △DCO and △ABO using coordinate geometry confirms that the ratio is 1 : 16. Thus the theoretical reasoning matches a coordinate based numeric check, which validates the result.

Why Other Options Are Wrong:

• 1 : 4 is incorrect because it uses the ratio of the bases directly rather than the squared ratio that arises from similar triangles.
• 1 : 2 is incorrect and does not match any standard proportional relationship in this configuration.
• 1 : 8 is also incorrect because it suggests a partial use of the base ratio but ignores the square relation between corresponding area and linear ratio.
• 4 : 1 reverses the smaller and larger triangle areas and does not agree with the geometry of how diagonals divide the trapezium.

Common Pitfalls:

• Confusing the ratio of sides AB : CD with the ratio of areas directly, without squaring.
• Mixing up which triangle has the larger area and inverting the ratio by mistake.
• Ignoring the fact that the diagonals are divided in the same ratio as the parallel sides, which is the crucial property for this problem.

Final Answer:
The ratio of the area of triangle DCO to the area of triangle ABO is 1 : 16.