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 $.
Let \( A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0. \) If \( B = P A P^T \), \( C = P^T B P \), and the sum of the diagonal elements of \( C \) is \( \frac{m}{n} \), where gcd(m, n) = 1, then \( m + n \) is: