ABCD is a trapezium in which AB is parallel to DC and 2AB = 3DC. The diagonals AC and BD intersect at O. What is the ratio of the area of Δ AOB to that of Δ DOC?
9 ∶ 4
Finding the Ratio of Areas of Triangles in a Trapezium
Let's analyze the given problem involving a trapezium ABCD. We are told that AB is parallel to DC, which is a key property of a trapezium. The diagonals AC and BD intersect at point O. We are also given a relationship between the lengths of the parallel sides: 2AB = 3DC.
We need to find the ratio of the area of triangle AOB to the area of triangle DOC, i.e., Area(Δ AOB) / Area(Δ DOC).
Identifying Similar Triangles
Consider the two triangles formed by the intersection of the diagonals and the parallel sides: Δ AOB and Δ DOC.
Since AB is parallel to DC and AC is a transversal, the alternate interior angles are equal:
- ∠ OAB = ∠ OCD (Alternate Interior Angles)
Similarly, since AB is parallel to DC and BD is a transversal, the alternate interior angles are equal:
- ∠ OBA = ∠ ODC (Alternate Interior Angles)
Also, the angles at the intersection point O are vertically opposite angles:
- ∠ AOB = ∠ DOC (Vertically Opposite Angles)
Because all three corresponding angles are equal, Δ AOB is similar to Δ DOC by the AAA (Angle-Angle-Angle) similarity criterion.
We can write this similarity as: Δ AOB ∼ Δ DOC.
Ratio of Areas of Similar Triangles
A fundamental property of similar triangles is that the ratio of their areas is equal to the square of the ratio of their corresponding sides. In our case, since Δ AOB ∼ Δ DOC, the corresponding sides are AB and DC, AO and DO, and BO and CO.
So, the ratio of their areas is:
\[ \frac{\text{Area}(\Delta \text{AOB})}{\text{Area}(\Delta \text{DOC})} = \left(\frac{\text{AB}}{\text{DC}}\right)^2 = \left(\frac{\text{AO}}{\text{CO}}\right)^2 = \left(\frac{\text{BO}}{\text{DO}}\right)^2 \]
Using the Given Relation between Parallel Sides
We are given the relation: \[ 2\text{AB} = 3\text{DC} \]
From this, we can find the ratio of the lengths of the parallel sides AB and DC:
\[ \frac{\text{AB}}{\text{DC}} = \frac{3}{2} \]
Calculating the Ratio of Areas
Now we can substitute the ratio AB/DC into the area ratio formula:
\[ \frac{\text{Area}(\Delta \text{AOB})}{\text{Area}(\Delta \text{DOC})} = \left(\frac{\text{AB}}{\text{DC}}\right)^2 = \left(\frac{3}{2}\right)^2 \]
Calculating the square:
\[ \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4} \]
So, the ratio of the area of Δ AOB to that of Δ DOC is 9:4.
Summary of Steps
- Identify the relevant triangles (Δ AOB and Δ DOC) formed by the diagonals intersecting in the trapezium.
- Prove that these two triangles are similar using angle properties (alternate interior angles due to parallel sides, vertically opposite angles).
- Use the property that the ratio of areas of similar triangles is the square of the ratio of corresponding sides.
- Use the given relation between the parallel sides (2AB = 3DC) to find the ratio AB/DC.
- Substitute this ratio into the area ratio formula and calculate the result.
Conclusion
The ratio of the area of Δ AOB to that of Δ DOC is 9 ∶ 4.
Revision Table: Trapezium Geometry
| Concept | Description | Application in Problem |
|---|---|---|
| Trapezium | A quadrilateral with at least one pair of parallel sides. | ABCD is a trapezium with AB || DC. |
| Parallel Lines & Transversals | When a transversal line intersects parallel lines, specific angle relationships exist (e.g., alternate interior angles are equal). | Diagonals AC and BD are transversals to parallel lines AB and DC, creating equal alternate interior angles (∠ OAB = ∠ OCD, ∠ OBA = ∠ ODC). |
| Vertically Opposite Angles | Angles opposite each other when two lines intersect are equal. | ∠ AOB and ∠ DOC are vertically opposite and thus equal. |
| Similar Triangles | Triangles with corresponding angles equal and corresponding sides proportional. | Δ AOB ∼ Δ DOC based on AAA similarity. |
| Ratio of Areas of Similar Triangles | The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. | Area(Δ AOB) / Area(Δ DOC) = (AB/DC)\(^2\). |
Additional Information: Properties of Trapeziums and Triangles
Trapeziums can have other specific properties. For example, an isosceles trapezium has non-parallel sides of equal length, and its diagonals are also equal in length. However, the problem only specifies a general trapezium with parallel sides AB and DC.
In any trapezium, the triangles formed by the non-parallel sides and the point of intersection of the diagonals (Δ AOD and Δ BOC) have equal areas. This is because Δ ABD and Δ ABC stand on the same base AB and are between the same parallel lines AB and DC, so Area(Δ ABD) = Area(Δ ABC). Subtracting Area(Δ AOB) from both gives Area(Δ AOD) = Area(Δ BOC).
Understanding the properties of similar triangles, especially the relationship between side ratios and area ratios, is crucial for solving geometry problems like this one.
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