ABCD is a trapezium in which AB is parallel to CD and AB = 4 (CD). The diagonals of the trapezium intersects at O. What is the ratio of area of triangle DCO to the area of the triangle ABO?

Options

A. `1:4`
B. `1:2`
C. `1:8`
D. `1:16`

Answer

To solve the problem, we need to find the ratio of the area of triangle DCO to the area of triangle ABO in trapezium ABCD, where AB is parallel to CD and AB = 4(CD).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let CD = x.
- Since AB = 4(CD), we have AB = 4x.
- The diagonals AC and BD intersect at point O.

2. **Use the Properties of Similar Triangles:**
- Since AB is parallel to CD, triangles ABO and DCO are similar by the AA (Angle-Angle) criterion. This is because:
- Angle A = Angle D (alternate interior angles)
- Angle B = Angle C (alternate interior angles)

3. **Set Up the Ratio of Areas:**
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
- Therefore, we can write:
\[ \frac{\text{Area of } \triangle ABO}{\text{Area of } \triangle DCO} = \left(\frac{AB}{CD}\right)^2 \]

4. **Substitute the Lengths:**
- We know AB = 4x and CD = x. Thus:
\[ \frac{AB}{CD} = \frac{4x}{x} = 4 \]
- Now, substituting this into the area ratio:
\[ \frac{\text{Area of } \triangle ABO}{\text{Area of } \triangle DCO} = 4^2 = 16 \]

5. **Find the Required Ratio:**
- We need the ratio of the area of triangle DCO to the area of triangle ABO:
\[ \frac{\text{Area of } \triangle DCO}{\text{Area of } \triangle ABO} = \frac{1}{16} \]

6. **Final Answer:**
- Therefore, the ratio of the area of triangle DCO to the area of triangle ABO is:
\[ \frac{\text{Area of } \triangle DCO}{\text{Area of } \triangle ABO} = \frac{1}{16} \]